Tangent and Normal Questions in English

(D) Given the curve equation: $\sqrt{x} + \sqrt{y} = 3$.
Differentiating both sides with respect to $x$:
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$.
Rearranging to find $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.
At the point $(4, 1)$,the slope $m$ is:
$m = \left. \frac{dy}{dx} \right|_{(4, 1)} = -\frac{\sqrt{1}}{\sqrt{4}} = -\frac{1}{2}$.
The length of the subtangent is given by the formula $\left| \frac{y}{dy/dx} \right|$.
Substituting the values $y = 1$ and $\frac{dy}{dx} = -\frac{1}{2}$:
$\text{Length of subtangent} = \left| \frac{1}{-1/2} \right| = |-2| = 2$.

(C) Let the curves touch at point $(x, y)$.
For the curves to touch,they must have the same value and the same derivative at the point of contact.
$1$. Equating the functions: $\frac{\ln x}{x} = \lambda x^2 \implies \ln x = \lambda x^3$.
$2$. Equating the derivatives: $\frac{d}{dx}(\frac{\ln x}{x}) = \frac{d}{dx}(\lambda x^2) \implies \frac{1 - \ln x}{x^2} = 2\lambda x \implies 1 - \ln x = 2\lambda x^3$.
Substitute $\ln x = \lambda x^3$ into the derivative equation:
$1 - (\lambda x^3) = 2\lambda x^3 \implies 1 = 3\lambda x^3 \implies \lambda x^3 = \frac{1}{3}$.
Since $\ln x = \lambda x^3$,we have $\ln x = \frac{1}{3}$,which implies $x = e^{1/3}$.
Now,substitute $x^3 = (e^{1/3})^3 = e$ into $\lambda x^3 = \frac{1}{3}$:
$\lambda e = \frac{1}{3} \implies \lambda = \frac{1}{3e}$.

(B) The curve is $f(x) = x^2 - 4x + 6 = (x-2)^2 + 2$. This is a parabola with vertex at $(2,2)$.
Let the line $L$ passing through $(2,3)$ have slope $m$. Its equation is $y - 3 = m(x - 2)$,or $y = mx - 2m + 3$.
The area bounded by the line and the parabola is minimized when the line is parallel to the tangent at the point where the chord is bisected. For a parabola,the chord passing through a point $(h, k)$ that bisects the area is the one where the slope of the chord is equal to the slope of the tangent at the point $(h, k)$ if the point lies on the curve,but here $(2,3)$ is inside the parabola.
Actually,for a parabola $y = ax^2 + bx + c$,the chord through a point $(x_0, y_0)$ that cuts off the minimum area is the one where the slope $m = f'(x_0)$.
Here $f'(x) = 2x - 4$. At $x = 2$,$f'(2) = 2(2) - 4 = 0$.
Thus,the slope of the line $L$ is $0$. The line $L$ is $y = 3$.
The tangent parallel to $L$ (slope $0$) is the horizontal tangent at the vertex $(2,2)$,which is $y = 2$.

(D) Given curve: $27x^2 = 4y^3$.
Differentiating with respect to $x$: $54x = 12y^2 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{9x}{2y^2}$.
Let the point on the curve be $(2t^3, 3t^2)$.
Slope of the tangent at $t$: $m = \frac{9(2t^3)}{2(3t^2)^2} = \frac{18t^3}{18t^4} = \frac{1}{t}$.
Equation of the tangent at $t$: $y - 3t^2 = \frac{1}{t}(x - 2t^3) \Rightarrow x - ty = -t^3$ .......$(i)$.
Equation of the normal at $t_1$: $y - 3t_1^2 = -t_1(x - 2t_1^3) \Rightarrow t_1x + y = 3t_1^2 + 2t_1^4$ .......$(ii)$.
Since the line is both tangent and normal,comparing $(i)$ and $(ii)$ gives $\frac{1}{t_1} = -t = \frac{-t^3}{3t_1^2 + 2t_1^4}$.
From $\frac{1}{t_1} = -t$,we have $t_1 = -\frac{1}{t}$.
Substituting into $-t = \frac{-t^3}{3t_1^2 + 2t_1^4}$: $t = \frac{t^3}{3(\frac{1}{t^2}) + 2(\frac{1}{t^4})} = \frac{t^7}{3t^2 + 2}$.
$3t^2 + 2 = t^6 \Rightarrow t^6 - 3t^2 - 2 = 0$. Let $u = t^2$,then $u^3 - 3u - 2 = 0$.
$(u - 2)(u + 1)^2 = 0$. Since $u = t^2 > 0$,we have $t^2 = 2$,so $t = \pm \sqrt{2}$.
Substituting $t^2 = 2$ into the tangent equation $x = t(y - t^2)$: $x = \pm \sqrt{2}(y - 2)$.

(B) Given the curve $y = x \sin x$.
At $x = \frac{\pi}{2}$,$y = \frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2} \times 1 = \frac{\pi}{2}$.
So,the point of contact is $(\frac{\pi}{2}, \frac{\pi}{2})$.
Now,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x \sin x) = x \cos x + \sin x$.
Calculate the slope $m$ at $x = \frac{\pi}{2}$:
$m = \left. \frac{dy}{dx} \right|_{x = \frac{\pi}{2}} = \frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) = \frac{\pi}{2}(0) + 1 = 1$.
The equation of the tangent is given by $y - y_1 = m(x - x_1)$:
$y - \frac{\pi}{2} = 1(x - \frac{\pi}{2})$.
$y - \frac{\pi}{2} = x - \frac{\pi}{2}$.
$x - y = 0$.

(B) Given the curve equation: $y^3 + 2xy + x^3 = (x - 1)^3$.
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} + 3x^2 = 3(x - 1)^2$.
At the point $(1, -1)$,substitute $x = 1$ and $y = -1$:
$3(-1)^2 \frac{dy}{dx} + 2(-1) + 2(1) \frac{dy}{dx} + 3(1)^2 = 3(1 - 1)^2$.
$3 \frac{dy}{dx} - 2 + 2 \frac{dy}{dx} + 3 = 0$.
$5 \frac{dy}{dx} + 1 = 0 \implies \frac{dy}{dx} = -\frac{1}{5}$.
The slope of the tangent at $(1, -1)$ is $m = -\frac{1}{5}$.
The slope of the normal is $m_{\perp} = -\frac{1}{m} = 5$.
The equation of the normal at $(1, -1)$ is:
$y - (-1) = 5(x - 1)$.
$y + 1 = 5x - 5$.
$5x - y = 6$.

(C) Given the curve equation is $e^y = 1 + x^2$.
Differentiating both sides with respect to $x$,we get:
$e^y \frac{dy}{dx} = 2x$
Thus,the slope of the tangent $m$ is given by:
$m = \frac{dy}{dx} = \frac{2x}{e^y} = \frac{2x}{1 + x^2}$
We need to find the range of $|m| = \left| \frac{2x}{1 + x^2} \right|$.
We know that for any real number $x$,$(|x| - 1)^2 \geq 0$.
$x^2 - 2|x| + 1 \geq 0$
$x^2 + 1 \geq 2|x|$
Dividing both sides by $x^2 + 1$ (which is always positive):
$1 \geq \frac{2|x|}{x^2 + 1}$
Therefore,$|m| = \left| \frac{2x}{1 + x^2} \right| \leq 1$.

(A) Given the curve $y = ax^3 + b$ and the point $A(2, 3)$ lies on the curve,we have:
$3 = a(2)^3 + b \implies 8a + b = 3$ (Equation $1$).
The slope of the tangent to the curve at any point $x$ is given by the derivative $\frac{dy}{dx} = 3ax^2$.
At $x = 2$,the slope is $m = 3a(2)^2 = 12a$.
The equation of the tangent is given as $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$.
Equating the slopes: $12a = 4 \implies a = \frac{4}{12} = \frac{1}{3}$.
Substitute $a = \frac{1}{3}$ into Equation $1$:
$8(\frac{1}{3}) + b = 3$
$\frac{8}{3} + b = 3$
$b = 3 - \frac{8}{3} = \frac{9 - 8}{3} = \frac{1}{3}$.
Thus,$b = \frac{1}{3}$.

(C) Given the curve $x = a(\cos t + \log \tan(t/2))$ and $y = a \sin t$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = a(-\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}) = a(-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}) = a(-\sin t + \frac{1}{\sin t}) = a(\frac{1-\sin^2 t}{\sin t}) = a \frac{\cos^2 t}{\sin t}$.
$\frac{dy}{dt} = a \cos t$.
Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = \tan t$.
The slope of the tangent $m_T = \tan t$. Let the angle of inclination be $\theta$,so $\tan \theta = \tan t$,which implies $\theta = t$.
The length of the tangent $PC$ is given by the formula $PC = |y \csc \theta|$.
Substituting $y = a \sin t$ and $\theta = t$:
$PC = |a \sin t \cdot \csc t| = |a \sin t \cdot \frac{1}{\sin t}| = |a|$.
Since $a$ is a constant length,the length of the tangent is $a$.

(A) The point $(1, 1)$ lies on the curve $xy + ax + by = 0$,so $1(1) + a(1) + b(1) = 0$,which gives $a + b = -1$ .......$(1)$
The slope of the tangent at $(1, 1)$ is given by $\frac{dy}{dx} = \tan(\tan^{-1}(2)) = 2$.
Differentiating the equation $xy + ax + by = 0$ with respect to $x$,we get:
$x \frac{dy}{dx} + y + a + b \frac{dy}{dx} = 0$
Substituting $x = 1$,$y = 1$,and $\frac{dy}{dx} = 2$ into the derivative equation:
$1(2) + 1 + a + b(2) = 0$
$3 + a + 2b = 0 \Rightarrow a + 2b = -3$ ........$(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(a + 2b) - (a + b) = -3 - (-1)$
$b = -2$
Substituting $b = -2$ into equation $(1)$:
$a - 2 = -1 \Rightarrow a = 1$
Finally,calculating the required value:
$\frac{a + b}{ab} = \frac{1 + (-2)}{1 \times (-2)} = \frac{-1}{-2} = \frac{1}{2}$

(C) The equation of the given line is $bx + ay = ab$,which can be rewritten as $y = -\frac{b}{a}x + b$. The slope of this line is $m = -\frac{b}{a}$.
The equation of the curve is $y = b \cdot e^{-x/a}$.
To find the slope of the tangent to the curve at any point $(x, y)$,we differentiate with respect to $x$:
$\frac{dy}{dx} = b \cdot e^{-x/a} \cdot \left(-\frac{1}{a}\right) = -\frac{b}{a} e^{-x/a}$.
Since the line touches the curve,the slope of the line must be equal to the slope of the tangent at the point of contact:
$-\frac{b}{a} = -\frac{b}{a} e^{-x/a}$.
Dividing both sides by $-\frac{b}{a}$ (assuming $a, b \neq 0$),we get:
$1 = e^{-x/a}$.
Since $e^0 = 1$,we have $-x/a = 0$,which implies $x = 0$.
Substituting $x = 0$ into the equation of the curve $y = b \cdot e^{-x/a}$:
$y = b \cdot e^0 = b \cdot 1 = b$.
Thus,the point of contact is $(0, b)$.

(B) Given the curve equation $2x^2 + y^2 = 12$.
Differentiating with respect to $x$,we get $4x + 2yy' = 0$,which implies $y' = -\frac{2x}{y}$.
At the point $(2, 2)$,the slope of the tangent $m_t = -\frac{2(2)}{2} = -2$.
The slope of the normal $m_n = -\frac{1}{m_t} = \frac{1}{2}$.
The equation of the normal at $(2, 2)$ is $y - 2 = \frac{1}{2}(x - 2)$,which simplifies to $2y - 4 = x - 2$,or $x = 2y - 2$.
Substituting $x = 2y - 2$ into the curve equation $2x^2 + y^2 = 12$:
$2(2y - 2)^2 + y^2 = 12$
$2(4y^2 - 8y + 4) + y^2 = 12$
$8y^2 - 16y + 8 + y^2 = 12$
$9y^2 - 16y - 4 = 0$
$(9y + 2)(y - 2) = 0$.
Thus,$y = 2$ or $y = -\frac{2}{9}$.
For $y = 2$,$x = 2(2) - 2 = 2$ (the original point).
For $y = -\frac{2}{9}$,$x = 2(-\frac{2}{9}) - 2 = -\frac{4}{9} - \frac{18}{9} = -\frac{22}{9}$.
Therefore,the normal meets the curve again at $\left( -\frac{22}{9}, -\frac{2}{9} \right)$.

(D) The length of the subtangent is given by $|y / (dy/dx)|$ and the length of the subnormal is given by $|y \cdot (dy/dx)|$.
Given that the length of the subnormal equals the length of the subtangent,we have $|y \cdot (dy/dx)| = |y / (dy/dx)|$,which implies $(dy/dx)^2 = 1$,so $dy/dx = \pm 1$.
Case $1$: If $dy/dx = 1$,the tangent at $(3, 4)$ is $y - 4 = 1(x - 3)$,which simplifies to $y = x + 1$. This line intersects the axes at $(0, 1)$ and $(-1, 0)$. Since the problem specifies the positive coordinate axes,this case is rejected.
Case $2$: If $dy/dx = -1$,the tangent at $(3, 4)$ is $y - 4 = -1(x - 3)$,which simplifies to $x + y = 7$. This line intersects the positive coordinate axes at $A(7, 0)$ and $B(0, 7)$.
The area of $\Delta OAB$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 7 = \frac{49}{2}$.

(D) Given $x = \sec^2 t$ and $y = \cot t$.
Since $\sec^2 t = 1 + \tan^2 t = 1 + \frac{1}{\cot^2 t} = 1 + \frac{1}{y^2}$,we have $x - 1 = \frac{1}{y^2}$,which implies $y^2(x - 1) = 1$.
At $t = \pi/4$,$x = \sec^2(\pi/4) = 2$ and $y = \cot(\pi/4) = 1$. So,$P = (2, 1)$.
Differentiating $y^2(x - 1) = 1$ with respect to $x$,we get $2y \frac{dy}{dx}(x - 1) + y^2 = 0$.
At $(2, 1)$,$2(1) \frac{dy}{dx}(2 - 1) + 1^2 = 0 \Rightarrow 2 \frac{dy}{dx} + 1 = 0 \Rightarrow \frac{dy}{dx} = -1/2$.
The equation of the tangent at $P(2, 1)$ is $y - 1 = -\frac{1}{2}(x - 2) \Rightarrow 2y - 2 = -x + 2 \Rightarrow x + 2y = 4$.
Substituting $y = \frac{4-x}{2}$ into the curve equation $y^2(x - 1) = 1$:
$\left(\frac{4-x}{2}\right)^2(x - 1) = 1 \Rightarrow (4-x)^2(x - 1) = 4 \Rightarrow (16 - 8x + x^2)(x - 1) = 4$.
$16x - 16 - 8x^2 + 8x + x^3 - x^2 = 4 \Rightarrow x^3 - 9x^2 + 24x - 20 = 0$.
Since $P(2, 1)$ is on the curve,$(x - 2)^2$ must be a factor. Dividing by $(x - 2)^2 = x^2 - 4x + 4$,we get $(x - 2)^2(x - 5) = 0$.
The roots are $x = 2$ (twice) and $x = 5$. Thus,the $x$-coordinate of $Q$ is $5$.

(B) Given the curve equation: $x^m y^n = a^{m+n}$.
Taking the natural logarithm on both sides: $m \ln x + n \ln y = \ln(a^{m+n})$.
Differentiating with respect to $x$: $\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{m}{n} \frac{y}{x}$.
The length of the subtangent is given by the formula: $LST = \left| \frac{y}{dy/dx} \right|$.
Substituting the value of $\frac{dy}{dx}$: $LST = \left| \frac{y}{(-\frac{m}{n} \frac{y}{x})} \right| = \left| -\frac{nx}{m} \right| = \frac{n}{m} |x|$.
Since $\frac{n}{m}$ is a constant,the length of the subtangent is proportional to the abscissa $x$.

(D) Given curve: $x^2y^2 - 2x = 4 - 4y$.
Differentiating with respect to $x$:
$2xy^2 + 2x^2y \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
At point $(2, -2)$:
$2(2)(-2)^2 + 2(2)^2(-2) \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
$16 - 16 \frac{dy}{dx} - 2 = -4 \frac{dy}{dx}$.
$14 = 12 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{14}{12} = \frac{7}{6}$.
Equation of the tangent at $(2, -2)$:
$y - (-2) = \frac{7}{6}(x - 2)$.
$6(y + 2) = 7(x - 2) \Rightarrow 6y + 12 = 7x - 14 \Rightarrow 7x - 6y = 26$.
Checking the options:
For $(-2, -7)$: $7(-2) - 6(-7) = -14 + 42 = 28 \neq 26$.
Thus,the tangent does not pass through $(-2, -7)$.

(A) Given the parametric equations of the curve:
$x = 2\cos t + 2t\sin t$
$y = 2\sin t - 2t\cos t$
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = -2\sin t + 2(\sin t + t\cos t) = 2t\cos t$
$\frac{dy}{dt} = 2\cos t - 2(\cos t - t\sin t) = 2t\sin t$
Now,find the slope of the tangent $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t\sin t}{2t\cos t} = \tan t$
At $t = \frac{\pi}{4}$,the slope of the tangent is $\tan(\frac{\pi}{4}) = 1$.
Therefore,the slope of the normal is $m = -\frac{1}{1} = -1$.
Find the point $(x, y)$ at $t = \frac{\pi}{4}$:
$x = 2\cos(\frac{\pi}{4}) + 2(\frac{\pi}{4})\sin(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) + \frac{\pi}{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} + \frac{\pi}{2\sqrt{2}}$
$y = 2\sin(\frac{\pi}{4}) - 2(\frac{\pi}{4})\cos(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) - \frac{\pi}{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} - \frac{\pi}{2\sqrt{2}}$
The equation of the normal is $y - y_1 = m(x - x_1)$:
$y - (\sqrt{2} - \frac{\pi}{2\sqrt{2}}) = -1(x - (\sqrt{2} + \frac{\pi}{2\sqrt{2}}))$
$y - \sqrt{2} + \frac{\pi}{2\sqrt{2}} = -x + \sqrt{2} + \frac{\pi}{2\sqrt{2}}$
$x + y = 2\sqrt{2}$
The distance of the line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$x + y - 2\sqrt{2} = 0$,so $A=1, B=1, C=-2\sqrt{2}$.
$d = \frac{|-2\sqrt{2}|}{\sqrt{1^2 + 1^2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2$.

(B) Given the curve $\sin y = x \sin \left( \frac{\pi}{3} + y \right)$.
At $x = 0$,$\sin y = 0 \implies y = 0$. So the point is $(0, 0)$.
Differentiating both sides with respect to $x$:
$\cos y \frac{dy}{dx} = \sin \left( \frac{\pi}{3} + y \right) + x \cos \left( \frac{\pi}{3} + y \right) \frac{dy}{dx}$.
At $(0, 0)$:
$\cos(0) \frac{dy}{dx} = \sin \left( \frac{\pi}{3} \right) + 0 \implies \frac{dy}{dx} = \frac{\sqrt{3}}{2}$.
The slope of the normal is $m_n = -\frac{1}{dy/dx} = -\frac{2}{\sqrt{3}}$.
The equation of the normal at $(0, 0)$ is $y - 0 = -\frac{2}{\sqrt{3}}(x - 0)$.
$\sqrt{3}y = -2x \implies 2x + \sqrt{3}y = 0$.

(D) Given the curve equation is $y = \cos(x + y)$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = -\sin(x + y) \left(1 + \frac{dy}{dx}\right) \dots (1)$
The equation of the tangent is $x + 2y = k$,which can be written as $y = -\frac{1}{2}x + \frac{k}{2}$.
The slope of the tangent is $m = -\frac{1}{2}$.
Substituting $\frac{dy}{dx} = -\frac{1}{2}$ into equation $(1)$:
$-\frac{1}{2} = -\sin(x + y) \left(1 - \frac{1}{2}\right)$
$-\frac{1}{2} = -\sin(x + y) \left(\frac{1}{2}\right)$
$\sin(x + y) = 1$
This implies $x + y = \frac{\pi}{2}$.
Substituting $x + y = \frac{\pi}{2}$ into the original curve equation $y = \cos(x + y)$:
$y = \cos\left(\frac{\pi}{2}\right) = 0$.
Since $y = 0$ and $x + y = \frac{\pi}{2}$,we have $x = \frac{\pi}{2}$.
Now,substitute the point $(\frac{\pi}{2}, 0)$ into the tangent equation $x + 2y = k$:
$\frac{\pi}{2} + 2(0) = k$
$k = \frac{\pi}{2}$.

(B) To find the intersection points,set the equations equal: $10 - x^2 = 2 + x^2$.
$2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$.
For $x = 2$,$y = 2 + (2)^2 = 6$. So,the intersection point is $(2, 6)$.
Find the slopes of the tangents at $(2, 6)$:
For $y = 10 - x^2$,$\frac{dy}{dx} = -2x$. At $x = 2$,$m_1 = -2(2) = -4$.
For $y = 2 + x^2$,$\frac{dy}{dx} = 2x$. At $x = 2$,$m_2 = 2(2) = 4$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$|\tan \theta | = |\frac{-4 - 4}{1 + (-4)(4)}| = |\frac{-8}{1 - 16}| = |\frac{-8}{-15}| = \frac{8}{15}$.

(B) Given the curve $y = xe^{x^2}$.
First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = x \cdot e^{x^2} \cdot (2x) + e^{x^2} \cdot 1 = e^{x^2}(2x^2 + 1)$.
At the point $(1, e)$,the slope $m$ is:
$m = e^{1^2}(2(1)^2 + 1) = e(2 + 1) = 3e$.
The equation of the tangent line at $(1, e)$ is given by $y - y_1 = m(x - x_1)$:
$y - e = 3e(x - 1)$
$y - e = 3ex - 3e$
$y = 3ex - 2e$.
Now,we check which point satisfies this equation $y = 3ex - 2e$:
For option $B$ $(\frac{4}{3}, 2e)$:
$y = 3e(\frac{4}{3}) - 2e = 4e - 2e = 2e$.
Since the point $(\frac{4}{3}, 2e)$ satisfies the equation,the tangent passes through this point.

(B) Given curve is $y = x^2 - 5x + 5$.
The slope of the tangent is given by $\frac{dy}{dx} = 2x - 5$.
The given line is $2y = 4x + 1$,which can be written as $y = 2x + \frac{1}{2}$. The slope of this line is $2$.
Since the tangent is parallel to the line,their slopes are equal: $2x - 5 = 2 \implies 2x = 7 \implies x = \frac{7}{2}$.
Substituting $x = \frac{7}{2}$ into the curve equation: $y = \left( \frac{7}{2} \right)^2 - 5\left( \frac{7}{2} \right) + 5 = \frac{49}{4} - \frac{35}{2} + 5 = \frac{49 - 70 + 20}{4} = -\frac{1}{4}$.
The point of tangency is $\left( \frac{7}{2}, -\frac{1}{4} \right)$.
The equation of the tangent line is $y - y_1 = m(x - x_1)$,where $m = 2$: $y - (-\frac{1}{4}) = 2(x - \frac{7}{2}) \implies y + \frac{1}{4} = 2x - 7 \implies y = 2x - 7 - \frac{1}{4} \implies y = 2x - \frac{29}{4}$.
Now,we check which point satisfies this equation. For option $B$: $x = \frac{1}{8}$,$y = 2(\frac{1}{8}) - \frac{29}{4} = \frac{1}{4} - \frac{29}{4} = -\frac{28}{4} = -7$.
Thus,the tangent passes through the point $\left( \frac{1}{8}, -7 \right)$.

(A) Given the curve $y = x^3 + ax - b$.
Since the point $(1, -5)$ lies on the curve,we have:
$-5 = (1)^3 + a(1) - b$
$-5 = 1 + a - b$
$a - b = -6$ $\dots(i)$
The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 3x^2 + a$.
At the point $(1, -5)$,the slope of the tangent is $m_1 = 3(1)^2 + a = 3 + a$.
The given line is $-x + y + 4 = 0$,which can be written as $y = x - 4$. The slope of this line is $m_2 = 1$.
Since the tangent is perpendicular to the line,the product of their slopes must be $-1$:
$m_1 \times m_2 = -1$
$(3 + a)(1) = -1$
$3 + a = -1$
$a = -4$.
Substituting $a = -4$ into equation $(i)$:
$-4 - b = -6$
$-b = -2$
$b = 2$.
Thus,the equation of the curve is $y = x^3 - 4x - 2$.
Now,we check the given options:
For $(2, -2)$: $y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2$.
Since the point $(2, -2)$ satisfies the equation,it lies on the curve.

(D) First,calculate the coordinates of the points $(1, f(1))$ and $(-1, f(-1))$.
$f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2$.
$f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0$.
The slope $m$ of the line segment joining $(1, -2)$ and $(-1, 0)$ is given by:
$m = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{-2 - 0}{2} = -1$.
The slope of the tangent to the curve $y = f(x)$ at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{d}{dx}(x^3 - x^2 - 2x) = 3x^2 - 2x - 2$.
Since the tangent is parallel to the line segment,their slopes must be equal:
$3x^2 - 2x - 2 = -1$.
$3x^2 - 2x - 1 = 0$.
Factoring the quadratic equation:
$3x^2 - 3x + x - 1 = 0$.
$3x(x - 1) + 1(x - 1) = 0$.
$(3x + 1)(x - 1) = 0$.
Thus,$x = 1$ or $x = -\frac{1}{3}$.
Therefore,$S = \left\{ -\frac{1}{3}, 1 \right\}$.

(B) Let the point $P$ be $(\alpha, \beta)$. Since $P$ lies on the curve,we have $\beta^{2}-3\alpha^{2}+\beta+10=0 \dots(i)$.
Differentiating the equation of the curve with respect to $x$,we get $2yy' - 6x + y' = 0$,which implies $y'(2y+1) = 6x$,so $y' = \frac{6x}{2y+1}$.
The slope of the tangent $m$ at $P(\alpha, \beta)$ is $m = \frac{6\alpha}{2\beta+1} \dots(ii)$.
The slope of the normal at $P$ is $-\frac{1}{m} = -\frac{2\beta+1}{6\alpha}$.
The normal passes through $(0, \frac{3}{2})$ and $(\alpha, \beta)$,so its slope is $\frac{\beta - 3/2}{\alpha - 0} = \frac{2\beta-3}{2\alpha}$.
Equating the slopes: $\frac{2\beta-3}{2\alpha} = -\frac{2\beta+1}{6\alpha}$.
Assuming $\alpha \neq 0$,we have $3(2\beta-3) = -(2\beta+1) \Rightarrow 6\beta - 9 = -2\beta - 1 \Rightarrow 8\beta = 8 \Rightarrow \beta = 1$.
Substituting $\beta = 1$ into $(i)$: $1^{2} - 3\alpha^{2} + 1 + 10 = 0 \Rightarrow 3\alpha^{2} = 12 \Rightarrow \alpha^{2} = 4$.
From $(ii)$,$|m| = |\frac{6\alpha}{2(1)+1}| = |\frac{6\alpha}{3}| = |2\alpha|$.
Since $\alpha^{2} = 4$,$|\alpha| = 2$,thus $|m| = 2 \times 2 = 4$.

(A) The slope of the tangent to the curve $y = f(x)$ at any point $x$ is given by the derivative $\frac{dy}{dx}$.
Given the curve $y = x^{3} - x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3x^{2} - 1$.
To find the slope of the tangent at $x = 2$,we substitute $x = 2$ into the derivative:
$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^{2} - 1 = 3(4) - 1 = 12 - 1 = 11$.
Therefore,the slope of the tangent at $x = 2$ is $11$.

(B) The slope of the tangent to the curve $y = \sqrt{4x - 3} - 1$ is given by the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x - 3} - 1) = \frac{1}{2\sqrt{4x - 3}} \times 4 = \frac{2}{\sqrt{4x - 3}}$.
Given that the slope is $\frac{2}{3}$,we set the derivative equal to $\frac{2}{3}$:
$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}$.
$\sqrt{4x - 3} = 3$.
Squaring both sides,we get $4x - 3 = 9$.
$4x = 12$,which implies $x = 3$.
Now,substitute $x = 3$ into the original equation to find $y$:
$y = \sqrt{4(3) - 3} - 1 = \sqrt{12 - 3} - 1 = \sqrt{9} - 1 = 3 - 1 = 2$.
Thus,the required point is $(3, 2)$.

(D) The given curve is $y = -\frac{2}{x-3}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x-3)^2}$.
Since the slope of the tangent is given as $2$,we have $\frac{2}{(x-3)^2} = 2$.
This implies $(x-3)^2 = 1$,so $x-3 = \pm 1$.
Thus,$x = 4$ or $x = 2$.
If $x = 4$,then $y = -\frac{2}{4-3} = -2$. The point is $(4, -2)$.
The equation of the tangent at $(4, -2)$ is $y - (-2) = 2(x - 4)$,which simplifies to $y + 2 = 2x - 8$ or $y - 2x + 10 = 0$.
If $x = 2$,then $y = -\frac{2}{2-3} = 2$. The point is $(2, 2)$.
The equation of the tangent at $(2, 2)$ is $y - 2 = 2(x - 2)$,which simplifies to $y - 2 = 2x - 4$ or $y - 2x + 2 = 0$.
Therefore,the equations of the tangents are $y - 2x + 2 = 0$ and $y - 2x + 10 = 0$.

(A) Given the equation of the curve: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$.
Differentiating both sides with respect to $x$,we get:
$\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$
$\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{25x}{4y}$.
The tangent is parallel to the $x$-axis if the slope $\frac{dy}{dx} = 0$.
Setting the slope to zero: $-\frac{25x}{4y} = 0$,which implies $x = 0$.
Substituting $x = 0$ into the original equation of the curve:
$\frac{0^{2}}{4} + \frac{y^{2}}{25} = 1$
$\frac{y^{2}}{25} = 1$
$y^{2} = 25$
$y = \pm 5$.
Thus,the points at which the tangents are parallel to the $x$-axis are $(0, 5)$ and $(0, -5)$.

(A) The given equation of the curve is $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$.
Differentiating both sides with respect to $x$,we get:
$\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$
$\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{25x}{4y}$
The tangent is parallel to the $y$-axis if the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator $y = 0$.
Substituting $y = 0$ into the equation of the curve:
$\frac{x^{2}}{4} + \frac{0^{2}}{25} = 1$
$\frac{x^{2}}{4} = 1 \implies x^{2} = 4 \implies x = \pm 2$.
Thus,the points are $(2, 0)$ and $(-2, 0)$.

(A) On the $x$-axis,the value of $y = 0$. Setting $y = 0$ in the curve equation $\frac{x-7}{(x-2)(x-3)} = 0$,we get $x = 7$. Thus,the point of intersection is $(7, 0)$.
To find the slope of the tangent,we differentiate $y = \frac{x-7}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x-7)(2x-5)}{(x^2 - 5x + 6)^2}$.
At the point $(7, 0)$,the denominator is $(7^2 - 5(7) + 6)^2 = (49 - 35 + 6)^2 = (20)^2 = 400$.
The numerator at $x=7$ is $(49 - 35 + 6) - (7-7)(2(7)-5) = 20 - 0 = 20$.
Thus,the slope $m = \frac{dy}{dx} \Big|_{(7,0)} = \frac{20}{400} = \frac{1}{20}$.
The equation of the tangent line at $(7, 0)$ with slope $m = \frac{1}{20}$ is given by $y - y_1 = m(x - x_1)$:
$y - 0 = \frac{1}{20}(x - 7)$,
$20y = x - 7$,
$20y - x + 7 = 0$.

(A) Given the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$.
Differentiating with respect to $x$,we get:
$\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} = 0$.
This simplifies to $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}$.
At the point $(1,1)$,the slope of the tangent $m_t = -\left(\frac{1}{1}\right)^{\frac{1}{3}} = -1$.
The equation of the tangent is $y - 1 = -1(x - 1)$,which simplifies to $x + y - 2 = 0$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal is $y - 1 = 1(x - 1)$,which simplifies to $y - x = 0$.

(A) The given curve is $y=3x^{4}-4x$.
The slope of the tangent to the curve at any point $x$ is given by the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{d}{dx}(3x^{4}-4x) = 12x^{3}-4$.
To find the slope of the tangent at $x=4$,we substitute $x=4$ into the derivative:
$\left(\frac{dy}{dx}\right)_{x=4} = 12(4)^{3}-4$.
$= 12(64)-4$.
$= 768-4 = 764$.
Thus,the slope of the tangent at $x=4$ is $764$.

(A) The given curve is $y = \frac{x-1}{x-2}$.
To find the slope of the tangent,we calculate the derivative $\frac{dy}{dx}$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x-2)\frac{d}{dx}(x-1) - (x-1)\frac{d}{dx}(x-2)}{(x-2)^2}$
$\frac{dy}{dx} = \frac{(x-2)(1) - (x-1)(1)}{(x-2)^2}$
$\frac{dy}{dx} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}$
Now,evaluate the slope at $x=10$:
$\left. \frac{dy}{dx} \right|_{x=10} = \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}$.
Thus,the slope of the tangent at $x=10$ is $\frac{-1}{64}$.

(A) The given curve is $y = x^{3} - x + 1$.
To find the slope of the tangent,we differentiate the equation with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^{3} - x + 1) = 3x^{2} - 1$.
The slope of the tangent at a point $(x_{0}, y_{0})$ is given by the value of $\frac{dy}{dx}$ at that point.
Given the $x$-coordinate $x_{0} = 2$,we substitute this into the derivative:
$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^{2} - 1$.
Calculating the value:
$3(4) - 1 = 12 - 1 = 11$.
Thus,the slope of the tangent at $x = 2$ is $11$.

(A) The given curve is $y = x^{3} - 3x + 2$.
To find the slope of the tangent,we differentiate the equation with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^{3} - 3x + 2) = 3x^{2} - 3$.
The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx}$.
We need to find the slope at the point where the $x$-coordinate is $3$.
Substituting $x = 3$ into the derivative:
$\left. \frac{dy}{dx} \right|_{x=3} = 3(3)^{2} - 3 = 3(9) - 3 = 27 - 3 = 24$.
Thus,the slope of the tangent at $x = 3$ is $24$.

(A) The equation of the given curve is $y=x^{3}-3x^{2}-9x+7$.
The slope of the tangent is given by the derivative $\frac{dy}{dx} = 3x^{2}-6x-9$.
The tangent is parallel to the $x$-axis if the slope is zero,i.e.,$\frac{dy}{dx} = 0$.
$3x^{2}-6x-9 = 0 \Rightarrow x^{2}-2x-3 = 0$.
$(x-3)(x+1) = 0$,which gives $x=3$ or $x=-1$.
For $x=3$,$y = (3)^{3}-3(3)^{2}-9(3)+7 = 27-27-27+7 = -20$.
For $x=-1$,$y = (-1)^{3}-3(-1)^{2}-9(-1)+7 = -1-3+9+7 = 12$.
Thus,the required points are $(3, -20)$ and $(-1, 12)$.

(A) If a tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$,then the slope of the tangent must be equal to the slope of the chord.
The slope of the chord joining $(x_1, y_1) = (2,0)$ and $(x_2, y_2) = (4,4)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} = 2$.
Now,the slope of the tangent to the curve $y = (x-2)^2$ at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 2(x-2)$.
Since the tangent is parallel to the chord,we equate the slopes: $2(x-2) = 2$.
Dividing by $2$,we get $x-2 = 1$,which implies $x = 3$.
To find the corresponding $y$-coordinate,substitute $x=3$ into the curve equation: $y = (3-2)^2 = 1^2 = 1$.
Thus,the required point on the curve is $(3,1)$.

(A) The equation of the given curve is $y=x^{3}-11x+5$.
The equation of the tangent is given as $y=x-11$,which is in the form $y=mx+c$.
Thus,the slope of the tangent is $m=1$.
The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 3x^{2}-11$.
Equating the slope of the tangent to the derivative,we get:
$3x^{2}-11 = 1$
$3x^{2} = 12$
$x^{2} = 4$
$x = \pm 2$.
When $x=2$,$y = (2)^{3}-11(2)+5 = 8-22+5 = -9$.
When $x=-2$,$y = (-2)^{3}-11(-2)+5 = -8+22+5 = 19$.
Therefore,the required points are $(2,-9)$ and $(-2,19)$.

(A) The equation of the given curve is $y=\frac{1}{x-1}, x \neq 1$.
The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = \frac{-1}{(x-1)^2}$.
Given that the slope of the tangent is $-1$,we set the derivative equal to $-1$:
$\frac{-1}{(x-1)^2} = -1$
$(x-1)^2 = 1$
$x-1 = \pm 1$
$x = 2$ or $x = 0$.
For $x=0$,$y = \frac{1}{0-1} = -1$. The point is $(0, -1)$.
For $x=2$,$y = \frac{1}{2-1} = 1$. The point is $(2, 1)$.
The equation of the tangent at $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
For point $(0, -1)$ and $m = -1$:
$y - (-1) = -1(x - 0) \Rightarrow y + 1 = -x \Rightarrow y + x + 1 = 0$.
For point $(2, 1)$ and $m = -1$:
$y - 1 = -1(x - 2) \Rightarrow y - 1 = -x + 2 \Rightarrow y + x - 3 = 0$.
Thus,the equations of the required lines are $y+x+1=0$ and $y+x-3=0$.

(A) The equation of the given curve is $y = \frac{1}{x-3}, x \neq 3$.
The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{d}{dx} (x-3)^{-1} = -1(x-3)^{-2} = \frac{-1}{(x-3)^2}$.
We are given that the slope of the tangent is $2$. Therefore,we set the derivative equal to $2$:
$\frac{-1}{(x-3)^2} = 2$.
This implies $(x-3)^2 = -\frac{1}{2}$.
Since the square of any real number $(x-3)^2$ must be non-negative,it cannot be equal to a negative value $-\frac{1}{2}$.
Thus,there is no real value of $x$ for which the slope of the tangent is $2$.
Hence,there are no tangents to the given curve with slope $2$.

(A) The equation of the given curve is $y=\frac{1}{x^{2}-2x+3}$.
The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.
Using the chain rule,$\frac{dy}{dx} = -\frac{1}{(x^{2}-2x+3)^{2}} \cdot (2x-2) = \frac{-2(x-1)}{(x^{2}-2x+3)^{2}}$.
Given that the slope of the tangent is $0$,we set $\frac{dy}{dx} = 0$:
$\frac{-2(x-1)}{(x^{2}-2x+3)^{2}} = 0$.
This implies $-2(x-1) = 0$,so $x = 1$.
Now,find the corresponding $y$-coordinate by substituting $x=1$ into the curve equation:
$y = \frac{1}{(1)^{2}-2(1)+3} = \frac{1}{1-2+3} = \frac{1}{2}$.
The point of tangency is $(1, \frac{1}{2})$.
The equation of a line with slope $m=0$ passing through $(x_0, y_0)$ is $y - y_0 = m(x - x_0)$.
Substituting the values,$y - \frac{1}{2} = 0(x - 1)$,which simplifies to $y = \frac{1}{2}$.

(A) The equation of the given curve is $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$.
Differentiating both sides with respect to $x$,we get:
$\frac{2x}{9} + \frac{2y}{16} \cdot \frac{dy}{dx} = 0$.
$\Rightarrow \frac{dy}{dx} = -\frac{16x}{9y}$.
The tangent is parallel to the $x$-axis if the slope of the tangent is $0$,i.e.,$\frac{dy}{dx} = 0$.
This implies $-\frac{16x}{9y} = 0$,which is possible only if $x = 0$.
Substituting $x = 0$ into the equation of the curve:
$\frac{0^{2}}{9} + \frac{y^{2}}{16} = 1
\Rightarrow y^{2} = 16
\Rightarrow y = \pm 4$.
Thus,the points on the curve where the tangents are parallel to the $x$-axis are $(0, 4)$ and $(0, -4)$.

(A) The equation of the given curve is $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$.
Differentiating both sides with respect to $x$,we get:
$\frac{2x}{9} + \frac{2y}{16} \cdot \frac{dy}{dx} = 0$.
This simplifies to $\frac{dy}{dx} = -\frac{16x}{9y}$.
The tangent is parallel to the $y$-axis when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator $y = 0$.
Substituting $y = 0$ into the equation of the curve:
$\frac{x^{2}}{9} + \frac{0^{2}}{16} = 1$
$\Rightarrow x^{2} = 9$
$\Rightarrow x = \pm 3$.
Thus,the points are $(3, 0)$ and $(-3, 0)$.

(A) Given the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 4x^{3}-18x^{2}+26x-10$.
At the point $(0,5)$,the slope of the tangent is:
$m = \left. \frac{dy}{dx} \right|_{(0,5)} = 4(0)^{3}-18(0)^{2}+26(0)-10 = -10$.
The equation of the tangent at $(0,5)$ is:
$y - 5 = -10(x - 0) \Rightarrow y - 5 = -10x \Rightarrow 10x + y - 5 = 0$.
The slope of the normal at $(0,5)$ is:
$m' = -\frac{1}{m} = -\frac{1}{-10} = \frac{1}{10}$.
The equation of the normal at $(0,5)$ is:
$y - 5 = \frac{1}{10}(x - 0) \Rightarrow 10y - 50 = x \Rightarrow x - 10y + 50 = 0$.

(A) Given the curve $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 4x^{3}-18x^{2}+26x-10$.
At the point $(1,3)$,the slope of the tangent is:
$m = \left. \frac{dy}{dx} \right|_{(1,3)} = 4(1)^{3}-18(1)^{2}+26(1)-10 = 4-18+26-10 = 2$.
The equation of the tangent at $(1,3)$ is:
$y-3 = 2(x-1) \Rightarrow y-3 = 2x-2 \Rightarrow 2x-y+1 = 0$.
The slope of the normal at $(1,3)$ is:
$m' = -\frac{1}{m} = -\frac{1}{2}$.
The equation of the normal at $(1,3)$ is:
$y-3 = -\frac{1}{2}(x-1) \Rightarrow 2y-6 = -x+1 \Rightarrow x+2y-7 = 0$.

(A) The equation of the curve is $y=x^{3}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 3x^{2}$.
The slope of the tangent at $(1,1)$ is $\left.\frac{dy}{dx}\right|_{(1,1)} = 3(1)^{2} = 3$.
The equation of the tangent at $(1,1)$ is $y-1 = 3(x-1)$,which simplifies to $3x-y-2=0$.
The slope of the normal at $(1,1)$ is $-\frac{1}{\text{slope of tangent}} = -\frac{1}{3}$.
The equation of the normal at $(1,1)$ is $y-1 = -\frac{1}{3}(x-1)$,which simplifies to $3y-3 = -x+1$,or $x+3y-4=0$.

(A) The equation of the curve is $y=x^{2}$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 2x$.
The slope of the tangent at $(0,0)$ is $\left. \frac{dy}{dx} \right|_{(0,0)} = 2(0) = 0$.
The equation of the tangent at $(0,0)$ is $y - 0 = 0(x - 0)$,which simplifies to $y = 0$.
The slope of the normal at $(0,0)$ is $\frac{-1}{\text{slope of tangent}} = \frac{-1}{0}$,which is undefined.
For a vertical line passing through $(x_0, y_0) = (0,0)$,the equation is $x = x_0$,so the equation of the normal is $x = 0$.

(A) The equation of the given curve is $y=x^{2}-2x+7$.
On differentiating with respect to $x$,we get:
$\frac{dy}{dx}=2x-2$.
The equation of the given line is $2x-y+9=0$,which can be written as $y=2x+9$.
This is in the form $y=mx+c$,so the slope of the line is $m=2$.
Since the tangent is parallel to the line $2x-y+9=0$,the slope of the tangent must be equal to the slope of the line.
Therefore,$2x-2=2$,which gives $2x=4$,so $x=2$.
Substituting $x=2$ into the curve equation,we get $y=(2)^{2}-2(2)+7=4-4+7=7$.
The point of tangency is $(2, 7)$.
The equation of the tangent line passing through $(2, 7)$ with slope $m=2$ is given by $y-y_{1}=m(x-x_{1})$.
$y-7=2(x-2)$
$y-7=2x-4$
$y-2x-3=0$.

(A) The equation of the line is $5y-15x=13$.
Rewriting in $y=mx+c$ form: $5y=15x+13 \Rightarrow y=3x+\frac{13}{5}$.
The slope of this line is $m_1=3$.
Since the tangent is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{3}$.
The slope of the tangent to the curve $y=x^2-2x+7$ is given by $\frac{dy}{dx} = 2x-2$.
Setting the slope equal to $-\frac{1}{3}$: $2x-2 = -\frac{1}{3} \Rightarrow 2x = 2 - \frac{1}{3} = \frac{5}{3} \Rightarrow x = \frac{5}{6}$.
Now,find the $y$-coordinate: $y = (\frac{5}{6})^2 - 2(\frac{5}{6}) + 7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{25-60+252}{36} = \frac{217}{36}$.
The point of tangency is $(\frac{5}{6}, \frac{217}{36})$.
The equation of the tangent line is $y - \frac{217}{36} = -\frac{1}{3}(x - \frac{5}{6})$.
Multiplying by $36$: $36y - 217 = -12(x - \frac{5}{6}) \Rightarrow 36y - 217 = -12x + 10$.
Rearranging gives $36y + 12x - 227 = 0$.