Tangent and Normal Questions in English

(N/A) The equation of the given curve is $y=7x^3+11$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 21x^2$.
The slope of the tangent to a curve at a point $(x_0, y_0)$ is given by $\left. \frac{dy}{dx} \right|_{(x_0, y_0)}$.
For $x=2$,the slope of the tangent is:
$m_1 = \left. \frac{dy}{dx} \right|_{x=2} = 21(2)^2 = 21 \times 4 = 84$.
For $x=-2$,the slope of the tangent is:
$m_2 = \left. \frac{dy}{dx} \right|_{x=-2} = 21(-2)^2 = 21 \times 4 = 84$.
Since the slopes of the tangents at $x=2$ and $x=-2$ are equal $(m_1 = m_2 = 84)$,the tangents are parallel.

(A) The equation of the given curve is $y=x^{3}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3x^{2}$.
The slope of the tangent at any point $(x, y)$ on the curve is given by $\frac{dy}{dx} = 3x^{2}$.
According to the problem,the slope of the tangent is equal to the $y$-coordinate of the point,so we have $3x^{2} = y$.
Since the point $(x, y)$ lies on the curve,we have $y = x^{3}$.
Equating the two expressions for $y$,we get $x^{3} = 3x^{2}$.
$x^{3} - 3x^{2} = 0$.
$x^{2}(x - 3) = 0$.
This gives $x = 0$ or $x = 3$.
If $x = 0$,then $y = (0)^{3} = 0$. So,the point is $(0, 0)$.
If $x = 3$,then $y = (3)^{3} = 27$. So,the point is $(3, 27)$.
Thus,the required points are $(0, 0)$ and $(3, 27)$.

(A) The equation of the curve is $y=4x^{3}-2x^{5}$.
The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = 12x^{2}-10x^{4}$.
The equation of the tangent at $(x, y)$ is $Y-y = (12x^{2}-10x^{4})(X-x)$.
Since the tangent passes through the origin $(0,0)$,we substitute $X=0$ and $Y=0$ into the equation:
$-y = (12x^{2}-10x^{4})(-x)$
$y = 12x^{3}-10x^{5}$.
Equating this to the original curve equation $y=4x^{3}-2x^{5}$:
$12x^{3}-10x^{5} = 4x^{3}-2x^{5}$
$8x^{3}-8x^{5} = 0$
$8x^{3}(1-x^{2}) = 0$.
This gives $x=0$ or $x^{2}=1$,so $x=0, 1, -1$.
For $x=0$,$y=4(0)^{3}-2(0)^{5} = 0$.
For $x=1$,$y=4(1)^{3}-2(1)^{5} = 2$.
For $x=-1$,$y=4(-1)^{3}-2(-1)^{5} = -2$.
Thus,the points are $(0,0), (1,2),$ and $(-1,-2)$.

(A) The equation of the given curve is $a y^{2}=x^{3}$.
Differentiating both sides with respect to $x$,we get:
$2 a y \frac{d y}{d x} = 3 x^{2}$
$\frac{d y}{d x} = \frac{3 x^{2}}{2 a y}$.
The slope of the tangent at the point $(a m^{2}, a m^{3})$ is:
$m_{t} = \left. \frac{d y}{d x} \right|_{(a m^{2}, a m^{3})} = \frac{3(a m^{2})^{2}}{2 a(a m^{3})} = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} = \frac{3 m}{2}$.
The slope of the normal at $(a m^{2}, a m^{3})$ is $m_{n} = -\frac{1}{m_{t}} = -\frac{2}{3 m}$.
The equation of the normal at $(a m^{2}, a m^{3})$ is given by:
$y - a m^{3} = -\frac{2}{3 m}(x - a m^{2})$
$3 m y - 3 a m^{4} = -2 x + 2 a m^{2}$
$2 x + 3 m y - 3 a m^{4} - 2 a m^{2} = 0$
$2 x + 3 m y - a m^{2}(2 + 3 m^{2}) = 0$.

(A) The equation of the given curve is $y=x^{3}+2x+6$.
The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} = 3x^{2}+2$.
The slope of the normal at any point $(x, y)$ is $-\frac{1}{\frac{dy}{dx}} = -\frac{1}{3x^{2}+2}$.
The equation of the given line is $x+14y+4=0$,which can be written as $y = -\frac{1}{14}x - \frac{4}{14}$.
The slope of this line is $-\frac{1}{14}$.
Since the normal is parallel to the line,their slopes must be equal:
$-\frac{1}{3x^{2}+2} = -\frac{1}{14} \Rightarrow 3x^{2}+2 = 14 \Rightarrow 3x^{2} = 12 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
For $x=2$,$y = (2)^{3} + 2(2) + 6 = 8 + 4 + 6 = 18$. The point is $(2, 18)$.
For $x=-2$,$y = (-2)^{3} + 2(-2) + 6 = -8 - 4 + 6 = -6$. The point is $(-2, -6)$.
The equation of the normal at $(2, 18)$ with slope $-\frac{1}{14}$ is:
$y - 18 = -\frac{1}{14}(x - 2) \Rightarrow 14y - 252 = -x + 2 \Rightarrow x + 14y - 254 = 0$.
The equation of the normal at $(-2, -6)$ with slope $-\frac{1}{14}$ is:
$y - (-6) = -\frac{1}{14}(x - (-2)) \Rightarrow y + 6 = -\frac{1}{14}(x + 2) \Rightarrow 14y + 84 = -x - 2 \Rightarrow x + 14y + 86 = 0$.
Thus,the equations are $x+14y-254=0$ and $x+14y+86=0$.

(N/A) The equations of the given curves are $x=y^{2}$ and $xy=k$.
Substituting $x=y^{2}$ into $xy=k$,we get:
$y^{2} \cdot y = k \Rightarrow y^{3} = k \Rightarrow y = k^{1/3}$.
Therefore,$x = (k^{1/3})^{2} = k^{2/3}$.
Thus,the point of intersection is $(k^{2/3}, k^{1/3})$.
Differentiating $x=y^{2}$ with respect to $x$,we have:
$1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.
The slope of the tangent $(m_{1})$ to $x=y^{2}$ at $(k^{2/3}, k^{1/3})$ is $m_{1} = \frac{1}{2k^{1/3}}$.
Differentiating $xy=k$ with respect to $x$,we have:
$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
The slope of the tangent $(m_{2})$ to $xy=k$ at $(k^{2/3}, k^{1/3})$ is $m_{2} = -\frac{k^{1/3}}{k^{2/3}} = -\frac{1}{k^{1/3}}$.
Two curves intersect at right angles if the product of their slopes is $-1$:
$m_{1} \cdot m_{2} = -1$
$\left(\frac{1}{2k^{1/3}}\right) \cdot \left(-\frac{1}{k^{1/3}}\right) = -1$
$-\frac{1}{2k^{2/3}} = -1$
$2k^{2/3} = 1$.
Cubing both sides,we get:
$(2k^{2/3})^{3} = 1^{3} \Rightarrow 8k^{2} = 1$.
Hence,the curves cut at right angles if $8k^{2}=1$.

(A) The equation of the given curve is $y=\sqrt{3x-2}$.
The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{3}{2\sqrt{3x-2}}$.
The equation of the given line is $4x-2y+5=0$,which can be written as $y=2x+\frac{5}{2}$.
The slope of this line is $m=2$.
Since the tangent is parallel to the line,their slopes must be equal:
$\frac{3}{2\sqrt{3x-2}} = 2$
$\Rightarrow \sqrt{3x-2} = \frac{3}{4}$
Squaring both sides,we get $3x-2 = \frac{9}{16}$.
$3x = \frac{9}{16} + 2 = \frac{41}{16}$
$x = \frac{41}{48}$.
Now,find the $y$-coordinate: $y = \sqrt{3(\frac{41}{48}) - 2} = \sqrt{\frac{41}{16} - 2} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
The point of tangency is $(\frac{41}{48}, \frac{3}{4})$.
The equation of the tangent is $y - \frac{3}{4} = 2(x - \frac{41}{48})$.
$y - \frac{3}{4} = 2x - \frac{41}{24}$.
Multiply by $24$: $24y - 18 = 48x - 41$.
$48x - 24y = 41 - 18 = 23$.
Thus,the equation is $48x-24y=23$.

(D) The equation of the given curve is $y=2x^{2}+3\sin x$.
First,we find the slope of the tangent to the curve by differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2x^{2}+3\sin x) = 4x + 3\cos x$.
The slope of the tangent at $x=0$ is:
$\left(\frac{dy}{dx}\right)_{x=0} = 4(0) + 3\cos(0) = 0 + 3(1) = 3$.
The slope of the normal is the negative reciprocal of the slope of the tangent:
$\text{Slope of normal} = -\frac{1}{\text{Slope of tangent}} = -\frac{1}{3}$.
Thus,the correct option is $D$.

(A) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y)(1 + \frac{dy}{dx})$.
Solving for $\frac{dy}{dx}$,we have $\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)}$.
The slope of the line $x + 2y = 0$ is $-\frac{1}{2}$.
Since the tangents are parallel to the line,$\frac{-\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2}$,which implies $2\sin(x + y) = 1 + \sin(x + y)$,so $\sin(x + y) = 1$.
Thus,$x + y = 2n\pi + \frac{\pi}{2}$ for $n \in \mathbb{Z}$.
Then $y = \cos(x + y) = \cos(2n\pi + \frac{\pi}{2}) = 0$.
Substituting $y = 0$ into $x + y = 2n\pi + \frac{\pi}{2}$,we get $x = 2n\pi + \frac{\pi}{2}$.
For $-2\pi \leq x \leq 2\pi$,the possible values for $x$ are $x = \frac{\pi}{2}$ (for $n=0$) and $x = -\frac{3\pi}{2}$ (for $n=-1$).
The points of tangency are $(\frac{\pi}{2}, 0)$ and $(-\frac{3\pi}{2}, 0)$.
The equation of the tangent at $(\frac{\pi}{2}, 0)$ is $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \Rightarrow 2x + 4y - \pi = 0$.
The equation of the tangent at $(-\frac{3\pi}{2}, 0)$ is $y - 0 = -\frac{1}{2}(x + \frac{3\pi}{2}) \Rightarrow 2x + 4y + 3\pi = 0$.

(A) Given the curve equations: $x=a \cos \theta+a \theta \sin \theta$ and $y=a \sin \theta-a \theta \cos \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -a \sin \theta + a \sin \theta + a \theta \cos \theta = a \theta \cos \theta$
$\frac{dy}{d\theta} = a \cos \theta - a \cos \theta + a \theta \sin \theta = a \theta \sin \theta$
Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$.
The slope of the normal is $-\frac{1}{dy/dx} = -\cot \theta = -\frac{\cos \theta}{\sin \theta}$.
The equation of the normal at point $(x, y)$ is $(Y - y) = -\cot \theta (X - x)$:
$Y - (a \sin \theta - a \theta \cos \theta) = -\frac{\cos \theta}{\sin \theta} (X - (a \cos \theta + a \theta \sin \theta))$
$Y \sin \theta - a \sin^2 \theta + a \theta \sin \theta \cos \theta = -X \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$
$X \cos \theta + Y \sin \theta = a(\cos^2 \theta + \sin^2 \theta) = a$
$X \cos \theta + Y \sin \theta - a = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $X \cos \theta + Y \sin \theta - a = 0$ is:
$d = \frac{|0 \cdot \cos \theta + 0 \cdot \sin \theta - a|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{|-a|}{\sqrt{1}} = |a|$.
Since $|a|$ is a constant,the distance of the normal from the origin is constant.

(B) The equation of the given curve is $2y + x^{2} = 3$.
Differentiating with respect to $x$,we get:
$2 \frac{dy}{dx} + 2x = 0$
$\frac{dy}{dx} = -x$
At the point $(1,1)$,the slope of the tangent is $\left. \frac{dy}{dx} \right|_{(1,1)} = -1$.
The slope of the normal at $(1,1)$ is given by $m_{normal} = \frac{-1}{\text{slope of tangent}} = \frac{-1}{-1} = 1$.
The equation of the normal at $(1,1)$ is $y - y_{1} = m_{normal}(x - x_{1})$.
$y - 1 = 1(x - 1)$
$y - 1 = x - 1$
$x - y = 0$.
Thus,the correct option is $B$.

(D) The equation of the given curve is $9y^{2} = x^{3}$.
Differentiating with respect to $x$,we get:
$9(2y) \frac{dy}{dx} = 3x^{2} \Rightarrow \frac{dy}{dx} = \frac{x^{2}}{6y}$.
The slope of the normal to the curve at point $(x_{1}, y_{1})$ is $-\frac{1}{\left( \frac{dy}{dx} \right)_{(x_{1}, y_{1})}} = -\frac{6y_{1}}{x_{1}^{2}}$.
The equation of the normal at $(x_{1}, y_{1})$ is $y - y_{1} = -\frac{6y_{1}}{x_{1}^{2}}(x - x_{1})$.
Rearranging into intercept form $\frac{x}{a} + \frac{y}{b} = 1$,we get:
$x_{1}^{2}y - x_{1}^{2}y_{1} = -6y_{1}x + 6x_{1}y_{1} \Rightarrow 6y_{1}x + x_{1}^{2}y = 6x_{1}y_{1} + x_{1}^{2}y_{1} = y_{1}(6x_{1} + x_{1}^{2})$.
Dividing by $y_{1}(6x_{1} + x_{1}^{2})$,we get the intercepts as $a = \frac{y_{1}(6x_{1} + x_{1}^{2})}{6y_{1}} = \frac{6x_{1} + x_{1}^{2}}{6}$ and $b = \frac{y_{1}(6x_{1} + x_{1}^{2})}{x_{1}^{2}}$.
Since the intercepts are equal,$a = b$ or $a = -b$. For the normal to make equal intercepts,the slope of the normal must be $-1$.
Thus,$-\frac{6y_{1}}{x_{1}^{2}} = -1 \Rightarrow x_{1}^{2} = 6y_{1}$.
Substituting $y_{1} = \frac{x_{1}^{2}}{6}$ into the curve equation $9y_{1}^{2} = x_{1}^{3}$:
$9 \left( \frac{x_{1}^{2}}{6} \right)^{2} = x_{1}^{3} \Rightarrow 9 \cdot \frac{x_{1}^{4}}{36} = x_{1}^{3} \Rightarrow \frac{x_{1}^{4}}{4} = x_{1}^{3}$.
Since $x_{1} \neq 0$,we have $x_{1} = 4$.
Then $y_{1}^{2} = \frac{4^{3}}{9} = \frac{64}{9} \Rightarrow y_{1} = \pm \frac{8}{3}$.
Thus,the points are $\left( 4, \pm \frac{8}{3} \right)$.

(A) Given curves are $y^{2}=x$ and $x^{2}=y$.
Solving these equations,we substitute $y=x^{2}$ into $y^{2}=x$,which gives $(x^{2})^{2}=x \Rightarrow x^{4}-x=0$.
$x(x^{3}-1)=0$,so $x=0$ or $x=1$.
For $x=0$,$y=0$ and for $x=1$,$y=1$. The points of intersection are $(0,0)$ and $(1,1)$.
For $y^{2}=x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx}=1$,so $\frac{dy}{dx}=\frac{1}{2y}$.
For $x^{2}=y$,differentiating with respect to $x$ gives $\frac{dy}{dx}=2x$.
At $(0,0)$,the slope of $y^{2}=x$ is undefined (vertical) and the slope of $x^{2}=y$ is $0$ (horizontal). Thus,the angle of intersection is $\frac{\pi}{2}$.
At $(1,1)$,the slope $m_{1}$ of $y^{2}=x$ is $\frac{1}{2(1)}=\frac{1}{2}$ and the slope $m_{2}$ of $x^{2}=y$ is $2(1)=2$.
The angle $\theta$ between them is given by $\tan \theta = |\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}| = |\frac{2-1/2}{1+(2)(1/2)}| = |\frac{3/2}{2}| = \frac{3}{4}$.
Thus,$\theta = \tan^{-1}(\frac{3}{4})$.
The angles of intersection are $\frac{\pi}{2}$ and $\tan^{-1}(\frac{3}{4})$.

(A) Let the curves intersect at $(x_{1}, y_{1})$.
For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,differentiating with respect to $x$ gives $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$,so $\frac{dy}{dx} = \frac{b^{2}x}{a^{2}y}$.
The slope of the tangent at $(x_{1}, y_{1})$ is $m_{1} = \frac{b^{2}x_{1}}{a^{2}y_{1}}$.
For the curve $xy=c^{2}$,differentiating with respect to $x$ gives $x\frac{dy}{dx}+y=0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
The slope of the tangent at $(x_{1}, y_{1})$ is $m_{2} = -\frac{y_{1}}{x_{1}}$.
For orthogonal intersection,$m_{1} \times m_{2} = -1$.
Substituting the slopes,we get $\left(\frac{b^{2}x_{1}}{a^{2}y_{1}}\right) \times \left(-\frac{y_{1}}{x_{1}}\right) = -1$.
This simplifies to $-\frac{b^{2}}{a^{2}} = -1$,which implies $a^{2} = b^{2}$ or $a^{2}-b^{2}=0$.

(A) Given the curve $y = \cos(x+y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x+y) \left(1 + \frac{dy}{dx}\right)$.
Rearranging,we have $\frac{dy}{dx} (1 + \sin(x+y)) = -\sin(x+y)$,so $\frac{dy}{dx} = -\frac{\sin(x+y)}{1 + \sin(x+y)}$.
Since the tangent is parallel to $x+2y=0$,its slope is $m = -\frac{1}{2}$.
Equating the derivative to the slope: $-\frac{\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2} \Rightarrow 2\sin(x+y) = 1 + \sin(x+y) \Rightarrow \sin(x+y) = 1$.
Since $\sin(x+y) = 1$,we have $\cos(x+y) = 0$. Given $y = \cos(x+y)$,this implies $y = 0$.
Substituting $y=0$ into the original equation: $0 = \cos(x+0) \Rightarrow \cos(x) = 0$.
For $-2\pi \leq x \leq 2\pi$,the solutions are $x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}$.
Checking $\sin(x+y) = 1$: For $x = \frac{\pi}{2}, y=0$,$\sin(\frac{\pi}{2}) = 1$ (Valid). For $x = -\frac{3\pi}{2}, y=0$,$\sin(-\frac{3\pi}{2}) = 1$ (Valid). For $x = -\frac{\pi}{2}, y=0$,$\sin(-\frac{\pi}{2}) = -1$ (Invalid). For $x = \frac{3\pi}{2}, y=0$,$\sin(\frac{3\pi}{2}) = -1$ (Invalid).
The points of tangency are $(\frac{\pi}{2}, 0)$ and $(-\frac{3\pi}{2}, 0)$.
Equation of tangent at $(\frac{\pi}{2}, 0)$ with slope $-\frac{1}{2}$: $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \Rightarrow 2x + 4y - \pi = 0$.
Equation of tangent at $(-\frac{3\pi}{2}, 0)$ with slope $-\frac{1}{2}$: $y - 0 = -\frac{1}{2}(x + \frac{3\pi}{2}) \Rightarrow 2x + 4y + 3\pi = 0$.

(N/A) Given the parametric equations of the curve:
$x = 3 \cos \theta - \cos^3 \theta$
$y = 3 \sin \theta - \sin^3 \theta$
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -3 \sin \theta + 3 \cos^2 \theta \sin \theta = -3 \sin \theta (1 - \cos^2 \theta) = -3 \sin^3 \theta$
$\frac{dy}{d\theta} = 3 \cos \theta - 3 \sin^2 \theta \cos \theta = 3 \cos \theta (1 - \sin^2 \theta) = 3 \cos^3 \theta$
Therefore,the slope of the tangent is:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3 \cos^3 \theta}{-3 \sin^3 \theta} = -\cot^3 \theta$
The slope of the normal is:
$m_n = -\frac{1}{dy/dx} = \frac{\sin^3 \theta}{\cos^3 \theta}$
The equation of the normal at point $(x, y)$ is:
$y - (3 \sin \theta - \sin^3 \theta) = \frac{\sin^3 \theta}{\cos^3 \theta} (x - (3 \cos \theta - \cos^3 \theta))$
Multiplying by $\cos^3 \theta$:
$y \cos^3 \theta - 3 \sin \theta \cos^3 \theta + \sin^3 \theta \cos^3 \theta = x \sin^3 \theta - 3 \cos \theta \sin^3 \theta + \sin^3 \theta \cos^3 \theta$
Simplifying:
$y \cos^3 \theta - x \sin^3 \theta = 3 \sin \theta \cos^3 \theta - 3 \cos \theta \sin^3 \theta$
$y \cos^3 \theta - x \sin^3 \theta = 3 \sin \theta \cos \theta (\cos^2 \theta - \sin^2 \theta)$
Using trigonometric identities $\sin 2\theta = 2 \sin \theta \cos \theta$ and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$:
$y \cos^3 \theta - x \sin^3 \theta = \frac{3}{2} \sin 2\theta \cos 2\theta = \frac{3}{4} \sin 4\theta$
Thus,$4(y \cos^3 \theta - x \sin^3 \theta) = 3 \sin 4\theta$.

(A) The curve is given by $y = b \cdot e^{-x / a}$.
To find the point where the curve intersects the $y$-axis,we set $x = 0$.
Substituting $x = 0$ into the curve equation,we get $y = b \cdot e^{0} = b \cdot 1 = b$.
Thus,the point of intersection is $(0, b)$.
Now,we find the slope of the tangent to the curve at $(0, b)$ by differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = b \cdot e^{-x / a} \cdot \left(-\frac{1}{a}\right) = -\frac{b}{a} e^{-x / a}$.
At the point $(0, b)$,the slope of the tangent is $\left(\frac{dy}{dx}\right)_{(0, b)} = -\frac{b}{a} e^{0} = -\frac{b}{a}$.
Next,we find the slope of the given line $\frac{x}{a} + \frac{y}{b} = 1$.
Rewriting the line equation in slope-intercept form $y = mx + c$,we have $\frac{y}{b} = -\frac{x}{a} + 1$,which implies $y = -\frac{b}{a}x + b$.
The slope of this line is $m = -\frac{b}{a}$.
Since the slope of the tangent to the curve at $(0, b)$ is equal to the slope of the line,and the point $(0, b)$ lies on both the curve and the line,the line touches the curve at the point where it intersects the $y$-axis.

(D) Given equation of the curve is $3x^{2}-y^{2}=8$ $(i)$.
On differentiating both sides with respect to $x$,we get $6x - 2y \frac{dy}{dx} = 0$.
$\Rightarrow \frac{dy}{dx} = \frac{6x}{2y} = \frac{3x}{y}$.
Slope of the tangent $m_{1} = \frac{3x}{y}$.
Slope of the normal $m_{2} = -\frac{1}{m_{1}} = -\frac{y}{3x}$ $(ii)$.
Since the normal is parallel to the line $x+3y=4$,its slope must be equal to the slope of the line.
For the line $x+3y=4$,$y = -\frac{1}{3}x + \frac{4}{3}$,so the slope $m_{3} = -\frac{1}{3}$.
Setting $m_{2} = m_{3}$,we get $-\frac{y}{3x} = -\frac{1}{3} \Rightarrow y = x$ $(iii)$.
Substituting $y=x$ into $(i)$,we get $3x^{2} - x^{2} = 8 \Rightarrow 2x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
If $x=2$,then $y=2$. If $x=-2$,then $y=-2$. The points are $(2, 2)$ and $(-2, -2)$.
The slope of the normal at these points is $m_{2} = -\frac{y}{3x} = -\frac{2}{3(2)} = -\frac{1}{3}$.
Equation of normal at $(2, 2)$ is $y - 2 = -\frac{1}{3}(x - 2) \Rightarrow 3y - 6 = -x + 2 \Rightarrow x + 3y = 8$.
Equation of normal at $(-2, -2)$ is $y + 2 = -\frac{1}{3}(x + 2) \Rightarrow 3y + 6 = -x - 2 \Rightarrow x + 3y = -8$.
Thus,the required equations are $x + 3y = \pm 8$.

(B) Given the curve $y(x)=\int_{0}^{x}(2t^{2}-15t+10)dt$.
By the Fundamental Theorem of Calculus,the slope of the tangent at $x=a$ is $y'(a) = 2a^{2}-15a+10$.
The slope of the normal at $(a, b)$ is $-\frac{1}{y'(a)} = -\frac{1}{2a^{2}-15a+10}$.
The given line is $x+3y=-5$,which can be written as $y=-\frac{1}{3}x-\frac{5}{3}$. The slope of this line is $-\frac{1}{3}$.
Since the normal is parallel to the line,their slopes are equal:
$-\frac{1}{2a^{2}-15a+10} = -\frac{1}{3} \Rightarrow 2a^{2}-15a+10 = 3$.
$2a^{2}-15a+7 = 0 \Rightarrow (2a-1)(a-7) = 0$.
Since $a>1$,we have $a=7$.
Now,calculate $b = y(7) = \int_{0}^{7}(2t^{2}-15t+10)dt = [\frac{2}{3}t^{3} - \frac{15}{2}t^{2} + 10t]_{0}^{7}$.
$b = \frac{2}{3}(343) - \frac{15}{2}(49) + 10(7) = \frac{686}{3} - \frac{735}{2} + 70 = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$.
Thus,$6b = -413$.
$|a+6b| = |7 - 413| = |-406| = 406$.

(D) Given,the equations of the curves are $2x = y^2 \dots(i)$ and $2xy = k \dots(ii)$.
From $(ii)$,$y = \frac{k}{2x}$. Substituting this into $(i)$,we get $2x = (\frac{k}{2x})^2 \Rightarrow 2x = \frac{k^2}{4x^2} \Rightarrow 8x^3 = k^2 \Rightarrow x = \frac{k^{2/3}}{2^{1/3} \cdot 2^{2/3}} = \frac{k^{2/3}}{2}$.
Then $y^2 = 2x = k^{2/3} \Rightarrow y = k^{1/3}$.
The point of intersection is $(\frac{k^{2/3}}{2}, k^{1/3})$.
Differentiating $(i)$ with respect to $x$: $2 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{y}$. Let $m_1 = \frac{1}{k^{1/3}}$.
Differentiating $(ii)$ with respect to $x$: $2y + 2x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
At the point of intersection,$m_2 = -\frac{k^{1/3}}{k^{2/3}/2} = -\frac{2}{k^{1/3}}$.
Since the curves intersect orthogonally,$m_1 \cdot m_2 = -1$.
$\frac{1}{k^{1/3}} \cdot (-\frac{2}{k^{1/3}}) = -1 \Rightarrow -\frac{2}{k^{2/3}} = -1 \Rightarrow k^{2/3} = 2$.
Cubing both sides,$k^2 = 8$.

(A) Given equations of the curves are $xy=4 \dots (i)$ and $x^{2}+y^{2}=8 \dots (ii)$.
Differentiating $(i)$ with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. Let this slope be $m_{1} = -\frac{y}{x}$.
Differentiating $(ii)$ with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. Let this slope be $m_{2} = -\frac{x}{y}$.
For the curves to touch,they must intersect and have the same slope at the point of intersection. Setting $m_{1} = m_{2}$ gives $-\frac{y}{x} = -\frac{x}{y} \Rightarrow y^{2} = x^{2}$.
Substituting $x^{2} = y^{2}$ into $(ii)$: $y^{2} + y^{2} = 8 \Rightarrow 2y^{2} = 8 \Rightarrow y^{2} = 4 \Rightarrow y = \pm 2$.
If $y = 2$,then $x = \frac{4}{2} = 2$. If $y = -2$,then $x = \frac{4}{-2} = -2$. The points of intersection are $(2, 2)$ and $(-2, -2)$.
At $(2, 2)$: $m_{1} = -\frac{2}{2} = -1$ and $m_{2} = -\frac{2}{2} = -1$. Since $m_{1} = m_{2}$,the curves touch at $(2, 2)$.
At $(-2, -2)$: $m_{1} = -\frac{-2}{-2} = -1$ and $m_{2} = -\frac{-2}{-2} = -1$. Since $m_{1} = m_{2}$,the curves touch at $(-2, -2)$.
Thus,the curves touch each other.

(A) Given the curve equation: $\sqrt{x}+\sqrt{y}=4$ $(i)$
Differentiating both sides with respect to $x$:
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$
Since the tangent is equally inclined to the axes,the slope of the tangent must be $\pm 1$. Thus,$\frac{dy}{dx} = \pm 1$.
Case $1$: $-\sqrt{\frac{y}{x}} = 1 \Rightarrow \sqrt{\frac{y}{x}} = -1$ (Not possible as square root is non-negative).
Case $2$: $-\sqrt{\frac{y}{x}} = -1 \Rightarrow \sqrt{\frac{y}{x}} = 1 \Rightarrow y = x$.
Substituting $y = x$ into equation $(i)$:
$\sqrt{x} + \sqrt{x} = 4$
$2\sqrt{x} = 4$
$\sqrt{x} = 2$
$x = 4$.
Since $y = x$,we have $y = 4$.
Therefore,the required coordinates are $(4, 4)$.

(A) Given curves are $y=4-x^{2} \dots(i)$ and $y=x^{2} \dots(ii)$.
To find the intersection points,equate the two equations: $x^{2} = 4 - x^{2} \Rightarrow 2x^{2} = 4 \Rightarrow x^{2} = 2 \Rightarrow x = \pm\sqrt{2}$.
For $x = \pm\sqrt{2}$,$y = 2$. Thus,the intersection points are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.
Now,find the slopes of the tangents at these points.
For curve $(i)$,$\frac{dy}{dx} = -2x$. At $x = \sqrt{2}$,$m_{1} = -2\sqrt{2}$.
For curve $(ii)$,$\frac{dy}{dx} = 2x$. At $x = \sqrt{2}$,$m_{2} = 2\sqrt{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 + (-2\sqrt{2})(2\sqrt{2})} \right| = \left| \frac{-4\sqrt{2}}{1 - 8} \right| = \left| \frac{-4\sqrt{2}}{-7} \right| = \frac{4\sqrt{2}}{7}$.
Therefore,$\theta = \tan^{-1}\left(\frac{4\sqrt{2}}{7}\right)$.

(C) The slope of the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$ is $m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$.
Since the tangent at $(a, b)$ is parallel to this line,the slope of the tangent $\left. \frac{dy}{dx} \right|_{(a, b)} = 1$.
Differentiating the curve $y = x + \sin y$ with respect to $x$,we get $\frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}$.
Substituting the point $(a, b)$ and the slope $1$,we have $1 = 1 + \cos b \cdot (1)$.
This implies $\cos b = 0$,which means $\sin b = \pm 1$.
Since $(a, b)$ lies on the curve,$b = a + \sin b$.
Therefore,$b - a = \sin b$.
Taking the absolute value on both sides,$|b - a| = |\sin b| = 1$.

(D) The point $P$ on the curve $y=x^{2}+7x+2$ nearest to the line $y=3x-3$ is the point where the tangent to the curve is parallel to the given line.
$1$. Find the slope of the tangent at any point $(x, y)$ on the curve:
$\frac{dy}{dx} = \frac{d}{dx}(x^{2}+7x+2) = 2x+7$
$2$. Since the tangent is parallel to the line $y=3x-3$,its slope must be equal to the slope of the line,which is $3$:
$2x+7 = 3$
$2x = -4$
$x = -2$
$3$. Find the $y$-coordinate of $P$ by substituting $x=-2$ into the curve equation:
$y = (-2)^{2} + 7(-2) + 2 = 4 - 14 + 2 = -8$
So,the point $P$ is $(-2, -8)$.
$4$. The normal at $P$ is perpendicular to the tangent. The slope of the tangent is $3$,so the slope of the normal is $-\frac{1}{3}$.
$5$. The equation of the normal at $P(-2, -8)$ with slope $m = -\frac{1}{3}$ is:
$y - (-8) = -\frac{1}{3}(x - (-2))$
$y + 8 = -\frac{1}{3}(x + 2)$
$3(y + 8) = -(x + 2)$
$3y + 24 = -x - 2$
$x + 3y + 26 = 0$

(B) For the curve $y=e^{x}$,the slope of the tangent at $(c, e^{c})$ is given by $\frac{dy}{dx} = e^{x} \implies m_{t} = e^{c}$.
The equation of the tangent at $(c, e^{c})$ is $y - e^{c} = e^{c}(x - c)$.
To find the intersection with the $x$-axis,set $y=0$: $-e^{c} = e^{c}(x - c) \implies -1 = x - c \implies x = c - 1$.
For the parabola $y^{2}=4x$,differentiating gives $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.
At the point $(1,2)$,the slope of the tangent is $m = \frac{2}{2} = 1$.
The slope of the normal is $m_{n} = -\frac{1}{m} = -1$.
The equation of the normal at $(1,2)$ is $y - 2 = -1(x - 1) \implies y - 2 = -x + 1 \implies x + y = 3$.
To find the intersection with the $x$-axis,set $y=0$: $x = 3$.
Since the points of intersection on the $x$-axis are the same,we equate the $x$-coordinates: $c - 1 = 3 \implies c = 4$.

(B) Given the curve equation: $x^{4} e^{y}+2 \sqrt{y+1}=3$.
Differentiating both sides with respect to $x$:
$x^{4} e^{y} y^{\prime} + 4x^{3} e^{y} + \frac{2 y^{\prime}}{2 \sqrt{y+1}} = 0$.
At the point $P(1,0)$,substitute $x=1$ and $y=0$:
$(1)^{4} e^{0} y^{\prime} + 4(1)^{3} e^{0} + \frac{y^{\prime}}{\sqrt{0+1}} = 0$.
$y^{\prime} + 4 + y^{\prime} = 0$.
$2y^{\prime} = -4$,which gives $y^{\prime} = -2$.
The equation of the tangent at $P(1,0)$ with slope $m = -2$ is:
$y - 0 = -2(x - 1)$.
$y = -2x + 2$,or $2x + y = 2$.
Checking the options:
For $(-2, 6)$,$2(-2) + 6 = -4 + 6 = 2$.
Thus,the point $(-2, 6)$ lies on the tangent line.

(D) The curve is given by $y = x^{2}-3x+2$.
To find the points where the curve intersects the $x$-axis,we set $y = 0$:
$x^{2}-3x+2 = 0$
$(x-1)(x-2) = 0$
So,the points of intersection are $A(1, 0)$ and $B(2, 0)$.
The slope of the tangent to the curve is given by $\frac{dy}{dx} = 2x - 3$.
At point $A(1, 0)$,the slope is $m_1 = 2(1) - 3 = -1$.
At point $B(2, 0)$,the slope is $m_2 = 2(2) - 3 = 1$.
The line $x+y=a$ has a slope of $-1$. Since it touches the curve at $A(1, 0)$,we substitute the coordinates into the line equation:
$1 + 0 = a \implies a = 1$.
The line $x-y=b$ has a slope of $1$. Since it touches the curve at $B(2, 0)$,we substitute the coordinates into the line equation:
$2 - 0 = b \implies b = 2$.
Therefore,$\frac{a}{b} = \frac{1}{2} = 0.50$.

(C) The slope of the line segment joining $(1, 0)$ and $(e, e)$ is given by $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$.
The derivative of the function $f(x) = x \log_{e} x$ is $f'(x) = \frac{d}{dx}(x) \cdot \log_{e} x + x \cdot \frac{d}{dx}(\log_{e} x) = 1 \cdot \log_{e} x + x \cdot \frac{1}{x} = \log_{e} x + 1$.
Since the tangent at $(c, f(c))$ is parallel to the line segment,its slope $f'(c)$ must be equal to $m$:
$f'(c) = \log_{e} c + 1 = \frac{e}{e - 1}$.
Solving for $\log_{e} c$:
$\log_{e} c = \frac{e}{e - 1} - 1 = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$.
Therefore,$c = e^{\frac{1}{e - 1}}$.

(A) The slope of the tangent to the curve $y=x^{3}$ at $P(t, t^{3})$ is given by $\frac{dy}{dx} = 3x^{2}$.
At $x=t$,the slope is $3t^{2}$.
The equation of the tangent at $P(t, t^{3})$ is $y - t^{3} = 3t^{2}(x - t)$.
To find the intersection with $y=x^{3}$,substitute $y=x^{3}$ into the tangent equation:
$x^{3} - t^{3} = 3t^{2}(x - t)$.
$(x - t)(x^{2} + xt + t^{2}) = 3t^{2}(x - t)$.
Since $x \neq t$ for point $Q$,we have $x^{2} + xt + t^{2} = 3t^{2}$,which simplifies to $x^{2} + xt - 2t^{2} = 0$.
Factoring gives $(x - t)(x + 2t) = 0$,so $x = -2t$.
The coordinates of $Q$ are $(-2t, (-2t)^{3}) = (-2t, -8t^{3})$.
The point dividing $PQ$ in the ratio $1:2$ has an ordinate given by the section formula:
$y = \frac{1 \times (-8t^{3}) + 2 \times (t^{3})}{1 + 2} = \frac{-8t^{3} + 2t^{3}}{3} = \frac{-6t^{3}}{3} = -2t^{3}$.

(C) Given the curve $y = ax^{2} + bx + c$.
Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation: $0 = a(0)^{2} + b(0) + c$,which gives $c = 0$.
The derivative of the curve is $\frac{dy}{dx} = 2ax + b$.
The tangent line at the origin $(0, 0)$ is $y = x$,which has a slope of $1$.
Thus,$\left. \frac{dy}{dx} \right|_{(0,0)} = 2a(0) + b = 1$,which gives $b = 1$.
Since the curve passes through the point $(1, 2)$,we substitute $x = 1, y = 2, b = 1, c = 0$ into the equation: $2 = a(1)^{2} + 1(1) + 0$.
This simplifies to $2 = a + 1$,so $a = 1$.
Therefore,the values are $a = 1, b = 1, c = 0$.

(C) Given curves are $x=y^{4}$ and $xy=k$.
For the intersection point,substitute $x=y^{4}$ into $xy=k$:
$y^{4} \cdot y = k \Rightarrow y^{5} = k \ldots(1)$.
Now,differentiate $x=y^{4}$ with respect to $x$:
$1 = 4y^{3} \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{4y^{3}}$.
Next,differentiate $xy=k$ with respect to $x$:
$y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
Since $x = \frac{k}{y}$,we have $\frac{dy}{dx} = -\frac{y}{k/y} = -\frac{y^{2}}{k}$.
Since the curves intersect at right angles,the product of their slopes is $-1$:
$\left(\frac{1}{4y^{3}}\right) \cdot \left(-\frac{y^{2}}{k}\right) = -1$.
$\Rightarrow \frac{1}{4yk} = 1 \Rightarrow y = \frac{1}{4k}$.
Substitute $y = \frac{1}{4k}$ into equation $(1)$:
$\left(\frac{1}{4k}\right)^{5} = k$.
$\Rightarrow \frac{1}{(4k)^{5}} = k$.
$\Rightarrow (4k)^{6} = 4$.

(D) Given the curve $y = x^{3} + 3x^{2} + 5$.
Let the point of tangency be $(x_{1}, y_{1})$.
The slope of the tangent at $(x_{1}, y_{1})$ is $\frac{dy}{dx} = 3x^{2} + 6x$.
So,the slope at $(x_{1}, y_{1})$ is $m = 3x_{1}^{2} + 6x_{1}$.
The equation of the tangent at $(x_{1}, y_{1})$ is $y - y_{1} = (3x_{1}^{2} + 6x_{1})(x - x_{1})$.
Since the tangent passes through the origin $(0, 0)$,we have:
$0 - y_{1} = (3x_{1}^{2} + 6x_{1})(0 - x_{1})$
$-y_{1} = -3x_{1}^{3} - 6x_{1}^{2} \implies y_{1} = 3x_{1}^{3} + 6x_{1}^{2} \quad (1)$.
Since $(x_{1}, y_{1})$ lies on the curve $y = x^{3} + 3x^{2} + 5$,we have:
$y_{1} = x_{1}^{3} + 3x_{1}^{2} + 5 \quad (2)$.
Equating $(1)$ and $(2)$:
$3x_{1}^{3} + 6x_{1}^{2} = x_{1}^{3} + 3x_{1}^{2} + 5$
$2x_{1}^{3} + 3x_{1}^{2} - 5 = 0$.
Testing for roots,$x_{1} = 1$ is a root.
Dividing by $(x_{1} - 1)$,we get $(x_{1} - 1)(2x_{1}^{2} + 5x_{1} + 5) = 0$.
For $x_{1} = 1$,$y_{1} = 3(1)^{3} + 6(1)^{2} = 9$.
Thus,the point is $(1, 9)$.
Checking which curve does not contain $(1, 9)$:
For $D$: $\frac{1}{3} - (9)^{2} = \frac{1}{3} - 81 \neq 2$.
Therefore,$(1, 9)$ does not lie on the curve $\frac{x}{3} - y^{2} = 2$.

(C) Given the parametric equations of the curve are $x = 12(t + \sin t \cos t)$ and $y = 12(1 + \sin t)^2$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 12(1 + \cos^2 t - \sin^2 t) = 12(1 + \cos 2t) = 12(2 \cos^2 t) = 24 \cos^2 t$.
$\frac{dy}{dt} = 12 \times 2(1 + \sin t) \cos t = 24(1 + \sin t) \cos t$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24(1 + \sin t) \cos t}{24 \cos^2 t} = \frac{1 + \sin t}{\cos t}$.
Since the angle made with the positive $x$-axis is $\frac{\pi}{3}$,the slope is $\tan(\frac{\pi}{3}) = \sqrt{3}$.
So,$\frac{1 + \sin t}{\cos t} = \sqrt{3} \Rightarrow 1 + \sin t = \sqrt{3} \cos t$.
Dividing by $2$,we get $\frac{1}{2} \sin t + \frac{\sqrt{3}}{2} \cos t = \frac{1}{2} \Rightarrow \sin(t + \frac{\pi}{3}) = \frac{1}{2}$.
Since $0 < t < \frac{\pi}{2}$,we have $\frac{\pi}{3} < t + \frac{\pi}{3} < \frac{5\pi}{6}$.
Thus,$t + \frac{\pi}{3} = \frac{5\pi}{6} \Rightarrow t = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.
Now,substitute $t = \frac{\pi}{6}$ into the expression for $y$:
$y_{0} = 12(1 + \sin(\frac{\pi}{6}))^2 = 12(1 + \frac{1}{2})^2 = 12(\frac{3}{2})^2 = 12 \times \frac{9}{4} = 27$.

(D) Given the curve equation: $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$.
To find the slope of the tangent at $(a, b)$,we differentiate with respect to $x$:
$n \left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} + n \left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \frac{dy}{dx} = 0$.
At the point $(a, b)$,we have $\frac{x}{a} = 1$ and $\frac{y}{b} = 1$.
Substituting these values: $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \frac{dy}{dx} = 0$.
$\frac{n}{a} + \frac{n}{b} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b}{a}$.
The equation of the tangent line at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$.
$y - b = -\frac{b}{a}x + b \implies \frac{b}{a}x + y = 2b$.
Dividing by $b$: $\frac{x}{a} + \frac{y}{b} = 2$.
This equation is independent of $n$ and holds for all $n \in N$.
Thus,$S = N$.

(A) The slope of the tangent to the curve $y=2x^2+x+2$ at any point $(h, k)$ is given by $\frac{dy}{dx} = 4x+1$.
At point $P(h, k)$,the slope of the tangent is $m_t = 4h+1$.
The slope of the normal line $\ell$ at $P$ is $m_n = -\frac{1}{4h+1}$.
The equation of the normal line $\ell$ passing through $P(h, k)$ is $y-k = -\frac{1}{4h+1}(x-h)$.
Since $Q(6, 4)$ lies on $\ell$,we have $4-k = -\frac{1}{4h+1}(6-h)$.
Substituting $k = 2h^2+h+2$,we get $4-(2h^2+h+2) = -\frac{6-h}{4h+1}$.
$(2-h-2h^2)(4h+1) = h-6$.
$8h+2-4h^2-h-8h^3-2h^2 = h-6$.
$-8h^3-6h^2+7h+2 = h-6$.
$8h^3+6h^2-6h-8 = 0$.
Dividing by $2$,$4h^3+3h^2-3h-4 = 0$.
$(h-1)(4h^2+7h+4) = 0$.
Since $4h^2+7h+4$ has no real roots (discriminant $D = 49-64 < 0$),we have $h=1$.
Then $k = 2(1)^2+1+2 = 5$. So $P$ is $(1, 5)$.
The area of $\triangle OPQ$ with vertices $O(0, 0)$,$P(1, 5)$,and $Q(6, 4)$ is given by:
Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(5-4) + 1(4-0) + 6(0-5)| = \frac{1}{2} |4 - 30| = \frac{1}{2} |-26| = 13$.

(D) Given the curve equation: $4x^{3}-3xy^{2}+6x^{2}-5xy-8y^{2}+9x+14=0$.
Differentiating with respect to $x$:
$12x^{2} - 3y^{2} - 6xyy' + 12x - 5y - 5xy' - 16yy' + 9 = 0$.
Substituting the point $(-2, 3)$:
$12(-2)^{2} - 3(3)^{2} - 6(-2)(3)y' + 12(-2) - 5(3) - 5(-2)y' - 16(3)y' + 9 = 0$.
$48 - 27 + 36y' - 24 - 15 + 10y' - 48y' + 9 = 0$.
$-9 - 2y' = 0 \implies y' = -\frac{9}{2}$.
Slope of tangent $m_{T} = -\frac{9}{2}$,slope of normal $m_{N} = \frac{2}{9}$.
Equation of tangent: $y - 3 = -\frac{9}{2}(x + 2) \implies y = -\frac{9}{2}x - 6$. For $y=0$,$x = -\frac{4}{3}$.
Equation of normal: $y - 3 = \frac{2}{9}(x + 2) \implies y = \frac{2}{9}x + \frac{31}{9}$. For $y=0$,$x = -\frac{31}{2}$.
Area $A = \frac{1}{2} \times |x_{T} - x_{N}| \times |y_{P}| = \frac{1}{2} \times |-\frac{4}{3} - (-\frac{31}{2})| \times 3 = \frac{3}{2} \times |-\frac{8}{6} + \frac{93}{6}| = \frac{3}{2} \times \frac{85}{6} = \frac{85}{4}$.
$8A = 8 \times \frac{85}{4} = 170$.

(C) Given that $A_{1} = 2A_{2}$.
From the graph,the total area of the rectangle formed by the coordinates $(x, y)$ is $A_{1} + A_{2} = xy - (4 \times 2) = xy - 8$.
Since $A_{1} = 2A_{2}$,we have $A_{1} + \frac{1}{2}A_{1} = xy - 8$,which implies $\frac{3}{2}A_{1} = xy - 8$,so $A_{1} = \frac{2}{3}xy - \frac{16}{3}$.
Also,$A_{1} = \int_{4}^{x} y \, dx$. Differentiating with respect to $x$ using the Leibniz rule:
$y = \frac{2}{3}(y + x \frac{dy}{dx}) \implies 3y = 2y + 2x \frac{dy}{dx} \implies y = 2x \frac{dy}{dx}$.
Separating variables: $\int \frac{dy}{y} = \int \frac{dx}{2x} \implies \ln y = \frac{1}{2} \ln x + C \implies y^2 = cx$.
Since the curve passes through $(4, 2)$,$2^2 = c(4) \implies c = 1$. Thus,$y^2 = x$.
The line is $2x - 12y = 15$,or $y = \frac{1}{6}x - \frac{15}{12}$. Its slope is $m = \frac{1}{6}$.
The normal is perpendicular to this line,so its slope is $m_n = -6$.
For $y^2 = x$,$2y \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2y}$.
Setting $\frac{dy}{dx} = -\frac{1}{m_n} = \frac{1}{6}$,we get $\frac{1}{2y} = \frac{1}{6} \implies y = 3$. Since $y^2 = x$,$x = 9$.
The point of contact is $(9, 3)$. The equation of the normal is $y - 3 = -6(x - 9) \implies y = -6x + 57$.
Checking the options: For $(10, 4)$,$4 = -6(10) + 57 = -3$,which is false. Thus,the normal does not pass through $(10, 4)$.

(B) Given the curve equation: $y^{5}-9xy+2x=0$.
Differentiating with respect to $x$:
$5y^{4}\frac{dy}{dx} - 9(y + x\frac{dy}{dx}) + 2 = 0$.
Rearranging for $\frac{dy}{dx}$:
$\frac{dy}{dx}(5y^{4} - 9x) = 9y - 2$.
$\frac{dy}{dx} = \frac{9y - 2}{5y^{4} - 9x}$.
For the tangent to be parallel to the $x$-axis,$\frac{dy}{dx} = 0$,which implies $9y - 2 = 0$,so $y = \frac{2}{9}$.
Substituting $y = \frac{2}{9}$ into the original equation:
$(\frac{2}{9})^{5} - 9x(\frac{2}{9}) + 2x = 0
\Rightarrow (\frac{2}{9})^{5} - 2x + 2x = 0
\Rightarrow (\frac{2}{9})^{5} = 0$,which is impossible.
Thus,there are no points where the tangent is parallel to the $x$-axis,so $M = 0$.
For the tangent to be parallel to the $y$-axis,the denominator $5y^{4} - 9x = 0$,so $x = \frac{5y^{4}}{9}$.
Substituting this into the original equation:
$y^{5} - 9y(\frac{5y^{4}}{9}) + 2(\frac{5y^{4}}{9}) = 0
\Rightarrow y^{5} - 5y^{5} + \frac{10y^{4}}{9} = 0
\Rightarrow -4y^{5} + \frac{10y^{4}}{9} = 0
\Rightarrow y^{4}(-4y + \frac{10}{9}) = 0$.
This gives $y = 0$ or $y = \frac{10}{36} = \frac{5}{18}$.
If $y = 0$,then $x = 0$. If $y = \frac{5}{18}$,then $x = \frac{5}{9}(\frac{5}{18})^{4}$.
Thus,there are $2$ points where the tangent is parallel to the $y$-axis,so $N = 2$.
Therefore,$M + N = 0 + 2 = 2$.

(C) The equation of the curve is $y=5x^{2}+2x-25$ at point $P(2, -1)$.
First,find the slope of the tangent at $P$ by differentiating: $y' = 10x + 2$.
At $x=2$,the slope $m = 10(2) + 2 = 22$.
The equation of the tangent line at $P(2, -1)$ is $y - (-1) = 22(x - 2)$,which simplifies to $y = 22x - 45$.
Now,consider the curve $y=x^{3}-x^{2}+x$ at point $(a, b)$.
The slope of the tangent at $(a, b)$ is given by $\frac{dy}{dx} = 3x^{2}-2x+1$.
At $x=a$,the slope is $3a^{2}-2a+1$.
Since this tangent is the same as the one found earlier,its slope must be $22$: $3a^{2}-2a+1 = 22$.
This gives the quadratic equation $3a^{2}-2a-21 = 0$.
Factoring the quadratic: $(3a+7)(a-3) = 0$,so $a=3$ or $a=-7/3$.
For $a=3$,the point $(a, b)$ lies on $y=x^{3}-x^{2}+x$,so $b = 3^{3}-3^{2}+3 = 27-9+3 = 21$.
Then $|2a+9b| = |2(3)+9(21)| = |6+189| = 195$.
For $a=-7/3$,the tangent line would be parallel but not identical to the line $y=22x-45$,so we reject this value.
Thus,the value is $195$.

(A) Given the curve $y = \frac{x-a}{(x+b)(x-2)}$.
Since the point $(1, -3)$ lies on the curve,we have $-3 = \frac{1-a}{(1+b)(1-2)}$.
$-3 = \frac{1-a}{-(1+b)} \implies 3(1+b) = 1-a \implies 1-a = 3+3b \implies a+3b = -2$ $(1)$.
The equation of the normal at $(1, -3)$ is $x - 4y = 13$,which can be written as $y = \frac{1}{4}x - \frac{13}{4}$.
The slope of the normal is $m_n = \frac{1}{4}$.
The slope of the tangent at $(1, -3)$ is $m_t = -\frac{1}{m_n} = -4$.
Now,differentiating $y = \frac{x-a}{x^2 + (b-2)x - 2b}$ with respect to $x$:
$\frac{dy}{dx} = \frac{(x^2 + (b-2)x - 2b)(1) - (x-a)(2x + b-2)}{(x^2 + (b-2)x - 2b)^2}$.
At $x=1$,$\frac{dy}{dx} = -4 = \frac{(1+b-2-2b) - (1-a)(2+b-2)}{(1+b-2-2b)^2} = \frac{-1-b - (1-a)b}{(-1-b)^2}$.
Since $1-a = 3(1+b)$,we substitute: $-4 = \frac{-(1+b) - 3(1+b)b}{(1+b)^2} = \frac{-(1+b)(1+3b)}{(1+b)^2} = \frac{-(1+3b)}{1+b}$.
$-4(1+b) = -1-3b \implies -4-4b = -1-3b \implies b = -3$.
Substituting $b = -3$ into $(1)$: $a + 3(-3) = -2 \implies a - 9 = -2 \implies a = 7$.
Therefore,$a+b = 7 + (-3) = 4$.

(C) The equation of the given line is $x+90y+2=0$,which can be written as $y=-\frac{1}{90}x-\frac{2}{90}$.
The slope of this line is $m=-\frac{1}{90}$.
Since the normal line is parallel to this line,the slope of the normal $(m_N)$ must be equal to $-\frac{1}{90}$.
The slope of the tangent $(m_T)$ is the negative reciprocal of the slope of the normal,so $m_T = -\frac{1}{m_N} = -\frac{1}{-1/90} = 90$.
We find the derivative of the curve $y=54x^5-135x^4-70x^3+180x^2+210x$:
$\frac{dy}{dx} = 270x^4 - 540x^3 - 210x^2 + 360x + 210$.
Setting the derivative equal to the slope of the tangent $(90)$:
$270x^4 - 540x^3 - 210x^2 + 360x + 210 = 90$
$270x^4 - 540x^3 - 210x^2 + 360x + 120 = 0$
Dividing by $30$:
$9x^4 - 18x^3 - 7x^2 + 12x + 4 = 0$.
Testing for roots,we find the roots are $x=1, x=2, x=-\frac{2}{3}, x=-\frac{1}{3}$.
Since there are $4$ distinct values of $x$,there are $4$ such points on the curve.

(C) Let the quadratic curve be $f(x) = A(x+1)(x-k)$. Since it touches $y=x$ at $(1,1)$,$f(1)=1$ and $f'(1)=1$.
$f(1) = A(2)(1-k) = 1 \Rightarrow 2A(1-k) = 1$.
$f'(x) = A(x-k) + A(x+1) = A(2x+1-k)$.
$f'(1) = A(2+1-k) = A(3-k) = 1$.
From $2A(1-k) = 1$ and $A(3-k) = 1$,we get $2A(1-k) = A(3-k) \Rightarrow 2-2k = 3-k \Rightarrow k = -1$.
Then $A(3 - (-1)) = 1 \Rightarrow 4A = 1 \Rightarrow A = 1/4$.
So,$f(x) = \frac{1}{4}(x+1)^2$.
Given the point $(\alpha, \alpha+1)$ lies on the curve: $\alpha+1 = \frac{1}{4}(\alpha+1)^2$.
Since $\alpha > -1$,$\alpha+1 = 4 \Rightarrow \alpha = 3$.
The point is $(3, 4)$.
$f'(x) = \frac{1}{2}(x+1)$,so $f'(3) = \frac{1}{2}(3+1) = 2$.
The slope of the normal at $(3, 4)$ is $m_n = -1/2$.
The equation of the normal is $y - 4 = -\frac{1}{2}(x - 3)$.
For the $x$-intercept,set $y=0$: $-4 = -\frac{1}{2}(x - 3) \Rightarrow 8 = x - 3 \Rightarrow x = 11$.

(A) The slope of the line joining the points $(c-1, e^{c-1})$ and $(c+1, e^{c+1})$ is given by $m = \frac{e^{c+1}-e^{c-1}}{(c+1)-(c-1)} = \frac{e^{c+1}-e^{c-1}}{2}$.
Since $e^x$ is a strictly convex function,the slope of the secant line between any two points is greater than the slope of the tangent at any point between them. Specifically,$\frac{e^{c+1}-e^{c-1}}{2} > e^c$ because $\frac{e-e^{-1}}{2} > 1$ (since $e \approx 2.718$,$\frac{2.718-0.368}{2} = 1.175 > 1$).
Because the slope of the secant line is greater than the slope of the tangent at $(c, e^c)$,and the function is convex,the tangent line must lie below the secant line for $x > c$ and above it for $x < c$. Therefore,the intersection point must occur to the left of $x=c$.
Alternatively,the equation of the tangent at $(c, e^c)$ is $y-e^c = e^c(x-c)$,or $y = e^c(x-c+1)$.
The equation of the secant line is $y-e^{c-1} = \frac{e^{c+1}-e^{c-1}}{2}(x-(c-1))$.
Solving these simultaneously for $x$ yields $x < c$.

(D) Given the curve equation: $(y-x^5)^2=x(1+x^2)^2$
Differentiating both sides with respect to $x$:
$2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + x \cdot 2(1+x^2)(2x)$
$2(y-x^5)(\frac{dy}{dx}-5x^4) = (1+x^2)^2 + 4x^2(1+x^2)$
Substitute the point $(1,3)$ into the equation:
$2(3-1^5)(\frac{dy}{dx}-5(1)^4) = (1+1^2)^2 + 4(1)^2(1+1^2)$
$2(3-1)(\frac{dy}{dx}-5) = (2)^2 + 4(2)$
$2(2)(\frac{dy}{dx}-5) = 4 + 8$
$4(\frac{dy}{dx}-5) = 12$
$\frac{dy}{dx}-5 = 3$
$\frac{dy}{dx} = 8$
Thus,the slope of the tangent at $(1,3)$ is $8$.

(A) Given curves are $xy=6$ $(1)$ and $x^2y=12$ $(2)$.
Dividing $(2)$ by $(1)$,we get $\frac{x^2y}{xy} = \frac{12}{6}$,which implies $x=2$.
Substituting $x=2$ in $(1)$,we get $2y=6$,so $y=3$. The point of intersection is $(2, 3)$.
For curve $(1)$,$y = \frac{6}{x}$,so $\frac{dy}{dx} = -\frac{6}{x^2}$. At $(2, 3)$,$m_1 = -\frac{6}{4} = -\frac{3}{2}$.
For curve $(2)$,$y = \frac{12}{x^2}$,so $\frac{dy}{dx} = -\frac{24}{x^3}$. At $(2, 3)$,$m_2 = -\frac{24}{8} = -3$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$\tan \theta = |\frac{-1.5 - (-3)}{1 + (-1.5)(-3)}| = |\frac{1.5}{1 + 4.5}| = |\frac{1.5}{5.5}| = \frac{15}{55} = \frac{3}{11}$.
Therefore,$\theta = \tan^{-1} \frac{3}{11}$.

(B) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y) \cdot (1 + \frac{dy}{dx})$.
Rearranging,$\frac{dy}{dx}(1 + \sin(x + y)) = -\sin(x + y)$,so $\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)}$.
The tangent is parallel to $x + 2y = 0$,which has a slope $m = -\frac{1}{2}$.
Setting $\frac{dy}{dx} = -\frac{1}{2}$,we get $\frac{-\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2} \implies 2\sin(x + y) = 1 + \sin(x + y) \implies \sin(x + y) = 1$.
Since $\sin(x + y) = 1$,then $x + y = 2n\pi + \frac{\pi}{2}$.
Substituting into the original equation: $y = \cos(2n\pi + \frac{\pi}{2}) = 0$.
Thus,$x + 0 = 2n\pi + \frac{\pi}{2} \implies x = 2n\pi + \frac{\pi}{2}$.
For $x \in [-2\pi, 2\pi]$,possible values are $x = \frac{\pi}{2}$ and $x = -\frac{3\pi}{2}$.
For $x = \frac{\pi}{2}, y = 0$,the tangent equation is $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \implies 2y = -x + \frac{\pi}{2} \implies 2x + 4y - \pi = 0$.
This matches option $B$.

(A) Given the curve equation is $y = ax^2 - 6x + b$.
Since the curve passes through $(0, 4)$,we substitute $x=0$ and $y=4$ into the equation:
$4 = a(0)^2 - 6(0) + b \implies b = 4$.
Now,the derivative of the curve is $\frac{dy}{dx} = 2ax - 6$.
The tangent is parallel to the $x$-axis at $x = \frac{3}{2}$,which means $\frac{dy}{dx} = 0$ at $x = \frac{3}{2}$.
$2a(\frac{3}{2}) - 6 = 0 \implies 3a - 6 = 0 \implies 3a = 6 \implies a = 2$.
Thus,the values are $a = 2$ and $b = 4$.

(A) Given the parametric equations of the curve: $x = 9(1 + \cos \theta)$ and $y = 9 \sin \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -9 \sin \theta$
$\frac{dy}{d\theta} = 9 \cos \theta$
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{9 \cos \theta}{-9 \sin \theta} = -\cot \theta$.
The slope of the normal is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{-\cot \theta} = \tan \theta = \frac{\sin \theta}{\cos \theta}$.
The equation of the normal at point $(9(1 + \cos \theta), 9 \sin \theta)$ is:
$y - 9 \sin \theta = \tan \theta (x - 9(1 + \cos \theta))$
$y - 9 \sin \theta = \frac{\sin \theta}{\cos \theta} (x - 9 - 9 \cos \theta)$
$y \cos \theta - 9 \sin \theta \cos \theta = x \sin \theta - 9 \sin \theta - 9 \sin \theta \cos \theta$
$y \cos \theta = x \sin \theta - 9 \sin \theta$
$y \cos \theta = \sin \theta (x - 9)$
If we check the point $(9, 0)$:
$0 \cdot \cos \theta = \sin \theta (9 - 9) \implies 0 = 0$.
Thus,the normal always passes through the fixed point $(9, 0)$.

(B) The equation of the curve is $xy = 1$,which can be written as $y = \frac{1}{x}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.
The slope of the tangent at any point $(x_1, y_1)$ on the curve is $m_t = -\frac{1}{x_1^2}$.
The slope of the normal at $(x_1, y_1)$ is $m_n = -\frac{1}{m_t} = x_1^2$.
The equation of the line is $ax + by + c = 0$,which can be written as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $-\frac{a}{b}$.
Since the line is normal to the curve,its slope must equal the slope of the normal: $-\frac{a}{b} = x_1^2$.
Since $x_1^2 > 0$ for all $x_1 \neq 0$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} < 0$.
This means $a$ and $b$ must have opposite signs.
Looking at the options,if $a > 0$,then $b < 0$ (Option $B$). If $a < 0$,then $b > 0$ (not explicitly listed as a single option,but $b < 0$ is given in $D$).
Given the standard form of such problems,$a$ and $b$ must have opposite signs,which is satisfied by $a > 0, b < 0$.