Tangent and Normal Questions in English

(C) Let the equation of the line parallel to the $x$-axis be $y = \lambda$.
Since the line intersects the curve $y = \sqrt{x}$,we have $\lambda = \sqrt{x}$,which implies $x = \lambda^2$.
Thus,the point of intersection is $P(\lambda^2, \lambda)$.
The slope of the curve $y = \sqrt{x}$ is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$.
At the point $P(\lambda^2, \lambda)$,the slope $m$ is $\frac{1}{2\sqrt{\lambda^2}} = \frac{1}{2|\lambda|}$.
The angle between the line (slope $0$) and the curve (slope $m$) is $45^\circ$.
Therefore,$\tan(45^\circ) = \left| \frac{m - 0}{1 + m \cdot 0} \right| = |m|$.
$1 = \left| \frac{1}{2\lambda} \right|$,which gives $2|\lambda| = 1$,so $\lambda = \pm \frac{1}{2}$.
Since $y = \sqrt{x}$ implies $y \ge 0$,we take $\lambda = \frac{1}{2}$.
Thus,the equation of the line is $y = \frac{1}{2}$.

(C) Given curves are $r_1 = \sin \theta + \cos \theta$ and $r_2 = 2 \sin \theta$.
At the point of intersection,$r_1 = r_2$,so $\sin \theta + \cos \theta = 2 \sin \theta$,which implies $\cos \theta = \sin \theta$,or $\tan \theta = 1$. Thus,$\theta = \frac{\pi}{4}$.
For $r_1 = \sin \theta + \cos \theta$,$\frac{dr_1}{d\theta} = \cos \theta - \sin \theta$. At $\theta = \frac{\pi}{4}$,$\frac{dr_1}{d\theta} = 0$. The angle $\phi_1$ is given by $\tan \phi_1 = \frac{r_1}{dr_1/d\theta} = \frac{\sqrt{2}}{0} = \infty$,so $\phi_1 = \frac{\pi}{2}$.
For $r_2 = 2 \sin \theta$,$\frac{dr_2}{d\theta} = 2 \cos \theta$. At $\theta = \frac{\pi}{4}$,$\frac{dr_2}{d\theta} = 2(\frac{1}{\sqrt{2}}) = \sqrt{2}$. The angle $\phi_2$ is given by $\tan \phi_2 = \frac{r_2}{dr_2/d\theta} = \frac{\sqrt{2}}{\sqrt{2}} = 1$,so $\phi_2 = \frac{\pi}{4}$.
The angle of intersection $\psi = |\phi_1 - \phi_2| = |\frac{\pi}{2} - \frac{\pi}{4}| = \frac{\pi}{4}$.

(B) The curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ intersect at $(0, 0)$ and $(\pi/2, 1)$.
Let the gradients of the tangents to the curves be $m_1$ and $m_2$ respectively.
For $y^2 = \frac{2x}{\pi}$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = \frac{2}{\pi}$,so $m_1 = \frac{dy}{dx} = \frac{1}{\pi y}$.
For $y = \sin x$,differentiating with respect to $x$ gives $m_2 = \frac{dy}{dx} = \cos x$.
At the point of intersection $(\pi/2, 1)$:
$m_1 = \frac{1}{\pi(1)} = \frac{1}{\pi}$.
$m_2 = \cos(\pi/2) = 0$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{1/\pi - 0}{1 + (1/\pi)(0)} \right| = \frac{1}{\pi}$.
Therefore,$\theta = \tan^{-1}(1/\pi) = \cot^{-1}(\pi)$.

(C) Given the curve $xy = a^2$,we have $y = \frac{a^2}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2}$.
At the point $(x_1, y_1)$,the slope of the tangent is $m = -\frac{a^2}{x_1^2}$.
The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{a^2}{x_1^2}(x - x_1)$.
Multiplying by $x_1^2$,we get $x_1^2 y - x_1^2 y_1 = -a^2 x + a^2 x_1$.
Since $x_1 y_1 = a^2$,this becomes $x_1^2 y + a^2 x = a^2 x_1 + x_1^2 y_1 = x_1(a^2 + x_1 y_1) = x_1(a^2 + a^2) = 2a^2 x_1$.
Dividing by $a^2 x_1$,the equation is $\frac{x}{2x_1} + \frac{y}{2a^2/x_1} = 1$.
This is the intercept form of a line $\frac{x}{a'} + \frac{y}{b'} = 1$,where the intercepts are $a' = 2x_1$ and $b' = \frac{2a^2}{x_1}$.
The area of the triangle formed by the axes and the tangent is $\frac{1}{2} \times |a'| \times |b'| = \frac{1}{2} \times (2x_1) \times \left(\frac{2a^2}{x_1}\right) = 2a^2$.

(B) Given the curve $y = \sin \left( \frac{\pi x}{2} \right)$.
First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at $(1, 1)$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \sin \left( \frac{\pi x}{2} \right) \right) = \cos \left( \frac{\pi x}{2} \right) \cdot \frac{\pi}{2}$.
Now,evaluate the slope of the tangent at $(1, 1)$:
$m_{tangent} = \left. \frac{dy}{dx} \right|_{(1, 1)} = \frac{\pi}{2} \cos \left( \frac{\pi(1)}{2} \right) = \frac{\pi}{2} \cos \left( \frac{\pi}{2} \right) = \frac{\pi}{2} \cdot 0 = 0$.
Since the slope of the tangent is $0$,the tangent is a horizontal line.
The slope of the normal is the negative reciprocal of the slope of the tangent,which is $-\frac{1}{0}$,representing a vertical line.
The equation of a vertical line passing through $(1, 1)$ is $x = 1$.

(C) Given the curve $y = 2 \cos x$.
At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2}$.
Now,find the derivative $\frac{dy}{dx} = \frac{d}{dx} (2 \cos x) = -2 \sin x$.
The slope of the tangent at $x = \frac{\pi}{4}$ is $m = \left( \frac{dy}{dx} \right)_{x = \pi/4} = -2 \sin \left( \frac{\pi}{4} \right) = -2 \left( \frac{1}{\sqrt{2}} \right) = -\sqrt{2}$.
The equation of the tangent line at point $(x_1, y_1) = \left( \frac{\pi}{4}, \sqrt{2} \right)$ with slope $m = -\sqrt{2}$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$.

(A) For the curve $y^2 = 4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$.
At the point $(1, 2)$,the slope $m_1 = \frac{2}{2} = 1$.
For the curve $x^2 + y^2 = 5$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.
At the point $(1, 2)$,the slope $m_2 = -\frac{1}{2}$.
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,$\tan \theta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} \right| = \left| \frac{3/2}{1/2} \right| = 3$.
Therefore,$\theta = \tan^{-1}(3)$.

(B) Given the curve $b{y^2} = {(x + a)^3}$.
Differentiating with respect to $x$,we get $2by \cdot \frac{dy}{dx} = 3{(x + a)^2}$.
Thus,$\frac{dy}{dx} = \frac{3{(x + a)^2}}{2by}$.
The formula for the subnormal is $y \cdot \frac{dy}{dx} = y \cdot \frac{3{(x + a)^2}}{2by} = \frac{3{(x + a)^2}}{2b}$.
The formula for the subtangent is $\frac{y}{\frac{dy}{dx}} = \frac{y}{\frac{3{(x + a)^2}}{2by}} = \frac{2by^2}{3{(x + a)^2}}$.
Substituting $y^2 = \frac{{(x + a)^3}}{b}$ into the subtangent expression:
Subtangent $= \frac{2b \cdot \frac{{(x + a)^3}}{b}}{3{(x + a)^2}} = \frac{2(x + a)}{3}$.
Therefore,$(Subtangent)^2 = \frac{4}{9}{(x + a)^2}$.
Now,consider the ratio $\frac{(Subtangent)^2}{Subnormal} = \frac{\frac{4}{9}{(x + a)^2}}{\frac{3}{2b}{(x + a)^2}} = \frac{8b}{27}$,which is a constant.
Hence,$(Subtangent)^2 \propto Subnormal$.

(C) Given the curve $y = ax^2 + bx$.
First,find the derivative $\frac{dy}{dx} = 2ax + b$.
The slope of the tangent at $(2, -8)$ is $\left( \frac{dy}{dx} \right)_{(2, -8)} = 2a(2) + b = 4a + b$.
Since the tangent is parallel to the $x$-axis,its slope must be $0$.
Therefore,$4a + b = 0 \Rightarrow b = -4a$ ..... $(i)$.
Since the point $(2, -8)$ lies on the curve,it must satisfy the equation $y = ax^2 + bx$.
Substituting $x = 2$ and $y = -8$: $-8 = a(2)^2 + b(2) \Rightarrow -8 = 4a + 2b$ ..... $(ii)$.
Substitute $b = -4a$ from $(i)$ into $(ii)$:
$-8 = 4a + 2(-4a)$
$-8 = 4a - 8a$
$-8 = -4a \Rightarrow a = 2$.
Now,find $b$ using $b = -4a$:
$b = -4(2) = -8$.
Thus,$a = 2$ and $b = -8$.

(A) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.
Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.
The equation of the tangent at point $(x_0, y_0)$ is $Y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}(X - x_0)$.
Simplifying,$X\sqrt{y_0} + Y\sqrt{x_0} = x_0\sqrt{y_0} + y_0\sqrt{x_0} = \sqrt{x_0y_0}(\sqrt{x_0} + \sqrt{y_0}) = \sqrt{x_0y_0}\sqrt{a}$.
Dividing by $\sqrt{a}\sqrt{x_0y_0}$,we get the intercept form: $\frac{X}{\sqrt{a}\sqrt{x_0}} + \frac{Y}{\sqrt{a}\sqrt{y_0}} = 1$.
The intercepts on the axes are $X_{int} = \sqrt{a}\sqrt{x_0}$ and $Y_{int} = \sqrt{a}\sqrt{y_0}$.
The sum of the intercepts is $\sqrt{a}(\sqrt{x_0} + \sqrt{y_0}) = \sqrt{a}(\sqrt{a}) = a$.

(D) Given the curve $y = x \log x$.
First,find the derivative $\frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = 1 + \log x$.
The slope of the tangent at any point $(x, y)$ is $m_t = 1 + \log x$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{1 + \log x}$.
The given line is $2x - 2y = 3$,which can be written as $y = x - \frac{3}{2}$. The slope of this line is $1$.
Since the normal is parallel to the line,their slopes must be equal:
$-\frac{1}{1 + \log x} = 1$
$-1 = 1 + \log x$
$\log x = -2$
$x = e^{-2}$.
Now,substitute $x = e^{-2}$ into the curve equation:
$y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$.
Thus,the required point is $(e^{-2}, -2e^{-2})$.

(C) The slope of the tangent to the curve $y = f(x)$ at any point is given by $\frac{dy}{dx}$.
The slope of the normal to the curve is given by $-\frac{1}{\frac{dy}{dx}}$.
Since the normal is parallel to the $x-$axis,its slope must be equal to $0$.
Therefore,$-\frac{1}{\frac{dy}{dx}} = 0$.
This implies that $\frac{dx}{dy} = 0$.

(C) The length of the normal is given by the formula: $L = |y| \sqrt{1 + (dy/dx)^2}$.
First,we find the derivatives with respect to $\theta$:
$dx/d\theta = a(1 + \cos \theta)$
$dy/d\theta = a \sin \theta$
Thus,$dy/dx = (dy/d\theta) / (dx/d\theta) = (a \sin \theta) / (a(1 + \cos \theta)) = \sin \theta / (1 + \cos \theta)$.
Using trigonometric identities,$\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$.
So,$dy/dx = \tan(\theta/2)$.
At $\theta = \pi/2$,$dy/dx = \tan(\pi/4) = 1$.
Also,at $\theta = \pi/2$,$y = a(1 - \cos(\pi/2)) = a(1 - 0) = a$.
Substituting these values into the formula:
$L = |a| \sqrt{1 + (1)^2} = a \sqrt{2}$.

(C) Given the curve: $x = a(\cos \theta + \theta \sin \theta )$ and $y = a(\sin \theta - \theta \cos \theta )$.
First,we find the derivatives with respect to $\theta$:
$\frac{dy}{d\theta} = a(\cos \theta - (\cos \theta - \theta \sin \theta)) = a\theta \sin \theta$.
$\frac{dx}{d\theta} = a(-\sin \theta + (\sin \theta + \theta \cos \theta)) = a\theta \cos \theta$.
Therefore,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 the negative reciprocal of the tangent slope:
$m_n = -\frac{1}{\tan \theta} = -\cot \theta = -\frac{\cos \theta}{\sin \theta}$.
The equation of the normal at point $(\theta)$ is:
$y - a(\sin \theta - \theta \cos \theta) = -\frac{\cos \theta}{\sin \theta} (x - a(\cos \theta + \theta \sin \theta))$.
Multiplying by $\sin \theta$:
$y \sin \theta - a \sin^2 \theta + a \theta \sin \theta \cos \theta = -x \cos \theta + a \cos^2 \theta + a \theta \sin \theta \cos \theta$.
Simplifying the equation:
$x \cos \theta + y \sin \theta = a(\sin^2 \theta + \cos^2 \theta) = a$.
The perpendicular distance from the origin $(0, 0)$ to the line $x \cos \theta + y \sin \theta - a = 0$ is:
$d = \frac{|-a|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |a| = a$.
Since $a$ is a constant,the normal is at a constant distance from the origin.

(A) Let the point of contact be $(h, k)$,where $k = h^4$.
The slope of the tangent to the curve $y = x^4$ at $(h, k)$ is given by $\frac{dy}{dx} = 4x^3$. At $x = h$,the slope is $m = 4h^3$.
The equation of the tangent at $(h, k)$ is $y - k = 4h^3(x - h)$.
Since the tangent passes through the point $(2, 0)$,we substitute these coordinates into the equation:
$0 - h^4 = 4h^3(2 - h)$
$-h^4 = 8h^3 - 4h^4$
$3h^4 - 8h^3 = 0$
$h^3(3h - 8) = 0$
This gives $h = 0$ or $h = \frac{8}{3}$.
For $h = 0$,the equation of the tangent is $y - 0 = 4(0)^3(x - 0)$,which simplifies to $y = 0$.
For $h = \frac{8}{3}$,the equation of the tangent is $y - (\frac{8}{3})^4 = 4(\frac{8}{3})^3(x - \frac{8}{3})$.
Thus,$y = 0$ is one of the equations of the tangent.

(D) Given curves are $y = x^2$ and $x = y^2$.
For $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.
At point $(1, 1)$,the slope $m_1 = 2(1) = 2$.
For $x = y^2$,differentiating with respect to $x$,we get $1 = 2y \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2y}$.
At point $(1, 1)$,the slope $m_2 = \frac{1}{2(1)} = \frac{1}{2}$.
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,$\tan \theta = \left| \frac{2 - 1/2}{1 + 2(1/2)} \right| = \left| \frac{3/2}{2} \right| = \frac{3}{4}$.
Therefore,$\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

(B) Given the curve equation is $y = 2x^2 - x + 1$.
To find the slope of the tangent,we differentiate the equation with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2x^2 - x + 1) = 4x - 1$.
The given line is $y = 3x + 9$,which is in the form $y = mx + c$,where the slope $m = 3$.
Since the tangent is parallel to the line $y = 3x + 9$,their slopes must be equal.
Therefore,$4x - 1 = 3$.
Solving for $x$:
$4x = 4 \implies x = 1$.
Now,substitute $x = 1$ into the original curve equation to find the $y$-coordinate:
$y = 2(1)^2 - (1) + 1 = 2 - 1 + 1 = 2$.
Thus,the required point is $(1, 2)$.

(C) Given the curve equation: ${x^3} - 8{a^2}y = 0$.
Differentiating with respect to $x$: $3{x^2} - 8{a^2} \frac{dy}{dx} = 0$.
Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{3{x^2}}{8{a^2}}$.
The slope of the normal is given by $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{8{a^2}}{3{x^2}}$.
We are given that the slope of the normal is $\frac{-2}{3}$.
Equating the two: $-\frac{8{a^2}}{3{x^2}} = -\frac{2}{3}$.
This simplifies to $\frac{8{a^2}}{{x^2}} = 2$,which means ${x^2} = 4{a^2}$,so $x = 2a$ (considering the positive coordinate).
Substituting $x = 2a$ into the original curve equation: ${(2a)^3} - 8{a^2}y = 0$.
$8{a^3} - 8{a^2}y = 0$,which implies $8{a^2}(a - y) = 0$,so $y = a$.
Therefore,the point is $(2a, a)$.

(C) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = a(1 + \cos t)$ and $\frac{dy}{dt} = a \sin t$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)} = \tan(t/2)$.
The length of the normal is given by the formula $L = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.
Substituting the values:
$L = |a(1 - \cos t)| \sqrt{1 + \tan^2(t/2)}$
$L = a(2 \sin^2(t/2)) \sqrt{\sec^2(t/2)}$
$L = 2a \sin^2(t/2) \sec(t/2)$
Since $\sec(t/2) = \frac{1}{\cos(t/2)}$,we have:
$L = 2a \sin(t/2) \cdot \frac{\sin(t/2)}{\cos(t/2)} = 2a \sin(t/2) \tan(t/2)$.

(B) Given the curve $y = e^{2x}$.
First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 2e^{2x}$.
At the point $(0, 1)$,the slope $m$ is: $m = \left. \frac{dy}{dx} \right|_{(0, 1)} = 2e^{2(0)} = 2(1) = 2$.
The equation of the tangent line at $(x_1, y_1) = (0, 1)$ with slope $m = 2$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 1 = 2(x - 0) \Rightarrow y = 2x + 1$.
To find where the tangent meets the $x-$axis,set $y = 0$:
$0 = 2x + 1 \Rightarrow 2x = -1 \Rightarrow x = -1/2$.
Thus,the tangent meets the $x-$axis at the point $(-1/2, 0)$.

(A) Given the curve equation: $(1 + x^2)y = 2 - x$ ......$(i)$
It crosses the $x$-axis where $y = 0$. Substituting $y = 0$ in $(i)$:
$0 = 2 - x \Rightarrow x = 2$.
So,the curve meets the $x$-axis at the point $(2, 0)$.
Now,express $y$ in terms of $x$:
$y = \frac{2 - x}{1 + x^2}$.
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(1 + x^2)(-1) - (2 - x)(2x)}{(1 + x^2)^2}$
$= \frac{-1 - x^2 - 4x + 2x^2}{(1 + x^2)^2} = \frac{x^2 - 4x - 1}{(1 + x^2)^2}$.
The slope of the tangent at $(2, 0)$ is:
$m = \left. \frac{dy}{dx} \right|_{(2, 0)} = \frac{2^2 - 4(2) - 1}{(1 + 2^2)^2} = \frac{4 - 8 - 1}{25} = \frac{-5}{25} = -\frac{1}{5}$.
The equation of the tangent at $(2, 0)$ with slope $m = -\frac{1}{5}$ is:
$y - 0 = -\frac{1}{5}(x - 2)$
$5y = -x + 2$
$x + 5y = 2$.

(C) Given curves are $y = x^2$ and $6y = 7 - x^3$.
For the curve $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.
At point $(1, 1)$,the slope $m_1 = 2(1) = 2$.
For the curve $6y = 7 - x^3$,differentiating with respect to $x$,we get $6 \frac{dy}{dx} = -3x^2$,which simplifies to $\frac{dy}{dx} = -\frac{x^2}{2}$.
At point $(1, 1)$,the slope $m_2 = -\frac{1^2}{2} = -\frac{1}{2}$.
Since $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the product of the slopes is $-1$.
Therefore,the curves intersect at a right angle,which is $\frac{\pi}{2}$.

(B) Given the curve equation is $y = 2x^2 - x + 1$.
The slope of the tangent at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 4x - 1$.
The given line is $y = 3x + 4$,which is in the slope-intercept form $y = mx + c$,where the slope $m = 3$.
Since the tangent is parallel to the line,their slopes must be equal. Therefore,we set the derivative equal to the slope of the line:
$4x - 1 = 3$
$4x = 4$
$x = 1$
Now,substitute $x = 1$ into the curve equation to find the $y$-coordinate of $P$:
$y = 2(1)^2 - (1) + 1$
$y = 2 - 1 + 1 = 2$
Thus,the coordinates of point $P$ are $(1, 2)$.

(D) Given the curve equation: $xy = c^2$ $(i)$
The formula for the subnormal at any point $(x, y)$ on a curve is given by: $\text{Subnormal} = y \frac{dy}{dx}$
From equation $(i)$,we have $y = \frac{c^2}{x}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -\frac{c^2}{x^2}$
Now,substitute $\frac{dy}{dx}$ into the subnormal formula:
$\text{Subnormal} = y \left( -\frac{c^2}{x^2} \right)$
Since $x = \frac{c^2}{y}$,substitute $x$ in the expression:
$\text{Subnormal} = y \left( -\frac{c^2}{(\frac{c^2}{y})^2} \right) = y \left( -\frac{c^2}{\frac{c^4}{y^2}} \right)$
$= y \left( -\frac{c^2 y^2}{c^4} \right) = -\frac{y^3}{c^2}$
Since $c$ is a constant,the magnitude of the subnormal varies as $y^3$.

(A) Given the curve equation is ${y^2} = 5x - 1$.
Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 5$,which implies $\frac{dy}{dx} = \frac{5}{2y}$.
At the point $(1, -2)$,the slope of the tangent is $m_t = \frac{5}{2(-2)} = -\frac{5}{4}$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{-5/4} = \frac{4}{5}$.
The equation of the normal at $(1, -2)$ is given by $(y - y_1) = m_n(x - x_1)$.
Substituting the values,we get $(y - (-2)) = \frac{4}{5}(x - 1)$.
$5(y + 2) = 4(x - 1) \implies 5y + 10 = 4x - 4$.
Rearranging the terms,we get $4x - 5y - 14 = 0$.
Comparing this with the given form $ax - 5y + b = 0$,we find $a = 4$ and $b = -14$.

(A) Given the curve equation is $y = 6x - x^2$ .....$(i)$
First,we find the slope of the tangent by differentiating the curve with respect to $x$:
$\frac{dy}{dx} = 6 - 2x$
The given line is $4x - 2y - 1 = 0$,which can be rewritten as $2y = 4x - 1$,or $y = 2x - 0.5$.
The slope of this line is $m = 2$.
Since the tangent is parallel to the line,their slopes must be equal:
$\frac{dy}{dx} = 2$
$6 - 2x = 2$
$2x = 4$
$x = 2$
Now,substitute $x = 2$ into the curve equation $(i)$ to find the $y$-coordinate:
$y = 6(2) - (2)^2 = 12 - 4 = 8$
Therefore,the point of tangency is $(2, 8)$.

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

(A) Given the parametric equations of the curve: $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(1 + \cos \theta)$
$\frac{dy}{d\theta} = a\sin \theta$
Now,calculate the slope of the tangent $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$
At $\theta = \frac{\pi}{2}$,$\frac{dy}{dx} = \frac{\sin(\pi/2)}{1 + \cos(\pi/2)} = \frac{1}{1 + 0} = 1$.
The value of $y$ at $\theta = \frac{\pi}{2}$ is $y = a(1 - \cos(\pi/2)) = a(1 - 0) = a$.
The length of the subtangent $ST$ is given by $\left| \frac{y}{dy/dx} \right| = \left| \frac{a}{1} \right| = a$.
The length of the subnormal $SN$ is given by $|y \cdot \frac{dy}{dx}| = |a \cdot 1| = a$.
Since $ST = a$ and $SN = a$,we conclude that $ST = SN$.

(B) Given the curve $x + y = e^{xy}$.
Differentiating both sides with respect to $x$:
$1 + \frac{dy}{dx} = e^{xy} \left( y + x \frac{dy}{dx} \right)$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$1 + \frac{dy}{dx} = y e^{xy} + x e^{xy} \frac{dy}{dx}$
$\frac{dy}{dx} (1 - x e^{xy}) = y e^{xy} - 1$
$\frac{dy}{dx} = \frac{y e^{xy} - 1}{1 - x e^{xy}}$
$A$ tangent is parallel to the $y-$ axis when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator is zero:
$1 - x e^{xy} = 0$
$x e^{xy} = 1$
Since $e^{xy} = x + y$,we substitute this into the equation:
$x(x + y) = 1$
$x^2 + xy = 1$
Checking the given options:
For $(1, 0)$: $1^2 + (1)(0) = 1 + 0 = 1$. This satisfies the condition.
Thus,the tangent is parallel to the $y-$ axis at the point $(1, 0)$.

(A) Given the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$.
Differentiating with respect to $x$,we get the slope of the tangent:
$\frac{dy}{dx} = 2x^2 + x$ ... $(i)$
Since the tangent makes an equal angle with the axes,the angle of inclination $\theta$ must be $45^\circ$ or $135^\circ$ (i.e.,$\pm 45^\circ$).
Thus,the slope $m = \tan(\pm 45^\circ) = \pm 1$.
Case $1$: $\frac{dy}{dx} = 1$
$2x^2 + x = 1 \implies 2x^2 + x - 1 = 0$
$(2x - 1)(x + 1) = 0 \implies x = \frac{1}{2}$ or $x = -1$.
Case $2$: $\frac{dy}{dx} = -1$
$2x^2 + x = -1 \implies 2x^2 + x + 1 = 0$.
The discriminant $D = 1^2 - 4(2)(1) = 1 - 8 = -7 < 0$,so there are no real solutions for this case.
Now,find the corresponding $y$-coordinates:
For $x = \frac{1}{2}$,$y = \frac{2}{3}(\frac{1}{8}) + \frac{1}{2}(\frac{1}{4}) = \frac{1}{12} + \frac{1}{8} = \frac{2+3}{24} = \frac{5}{24}$.
For $x = -1$,$y = \frac{2}{3}(-1)^3 + \frac{1}{2}(-1)^2 = -\frac{2}{3} + \frac{1}{2} = \frac{-4+3}{6} = -\frac{1}{6}$.
Therefore,the required points are $\left( \frac{1}{2}, \frac{5}{24} \right)$ and $\left( -1, -\frac{1}{6} \right)$.

(D) Given the curve equation: $y^3 + 3x^2 = 12y$.
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$.
Rearranging to find $\frac{dy}{dx}$:
$\frac{dy}{dx}(3y^2 - 12) = -6x \implies \frac{dy}{dx} = \frac{-6x}{3y^2 - 12} = \frac{2x}{4 - y^2}$.
For the tangent to be vertical (parallel to the $y$-axis),the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero:
$4 - y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$.
Case $1$: If $y = 2$,then $2^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}}$.
Case $2$: If $y = -2$,then $(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$.
Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

(D) Given the curve is $y^2 = x$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,which implies $\frac{dy}{dx} = \frac{1}{2y}$.
The slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{1}{2y_1}$.
Since the tangent makes an angle of $45^{\circ}$ with the $x-$axis,the slope $m = \tan(45^{\circ}) = 1$.
Equating the slopes,$\frac{1}{2y_1} = 1$,which gives $y_1 = \frac{1}{2}$.
Substituting $y_1 = \frac{1}{2}$ into the curve equation $y^2 = x$,we get $x_1 = (\frac{1}{2})^2 = \frac{1}{4}$.
Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

(A) Given the equation of the parabola: $y = 2 + 4x - 4x^2$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2 + 4x - 4x^2) = 4 - 8x$.
We are given that the slope of the tangent is $-4$.
So,$4 - 8x = -4$.
$-8x = -8 \Rightarrow x = 1$.
Now,find the corresponding $y$-coordinate by substituting $x = 1$ into the original equation:
$y = 2 + 4(1) - 4(1)^2 = 2 + 4 - 4 = 2$.
The point of tangency is $(1, 2)$.
The equation of the tangent line with slope $m = -4$ passing through $(1, 2)$ is given by:
$y - y_1 = m(x - x_1)$
$y - 2 = -4(x - 1)$
$y - 2 = -4x + 4$
$4x + y - 6 = 0$.

(D) Given curve is $2y = 3 - x^2$.
Differentiating with respect to $x$,we get $2 \frac{dy}{dx} = -2x$,which implies $\frac{dy}{dx} = -x$.
The slope of the tangent at $(1, 1)$ is $m = \left(\frac{dy}{dx}\right)_{(1,1)} = -1$.
The slope of the normal at $(1, 1)$ is $-\frac{1}{m} = -\frac{1}{-1} = 1$.
The equation of the normal at $(1, 1)$ is $(y - 1) = 1(x - 1)$.
Simplifying this,we get $y - 1 = x - 1$,which results in $x - y = 0$.

(B) The given curve is $y = -\frac{2}{x - 3}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = \frac{2}{(x - 3)^2}$.
Given that the slope of the tangent is $2$,we set $\frac{2}{(x - 3)^2} = 2$.
This simplifies to $(x - 3)^2 = 1$,which means $x - 3 = \pm 1$.
Thus,$x = 4$ or $x = 2$.
For $x = 4$,$y = -\frac{2}{4 - 3} = -2$. The point is $(4, -2)$.
For $x = 2$,$y = -\frac{2}{2 - 3} = 2$. The point is $(2, 2)$.
The equation of the tangent at $(4, -2)$ with slope $2$ is $y - (-2) = 2(x - 4)$,which simplifies to $y + 2 = 2x - 8$ or $y - 2x + 10 = 0$.
The equation of the tangent at $(2, 2)$ with slope $2$ is $y - 2 = 2(x - 2)$,which simplifies to $y - 2 = 2x - 4$ or $y - 2x + 2 = 0$.
Comparing with the given options,$y - 2x + 10 = 0$ is the correct choice.

(D) Let the point be $(x_1, y_1)$.
Given the curve $y = x^3 + 5$.
The slope of the tangent is $\frac{dy}{dx} = 3x^2$.
At point $(x_1, y_1)$,the slope is $m_1 = 3x_1^2$.
The given line is $x + 3y = 2$,which can be written as $3y = -x + 2$ or $y = -\frac{1}{3}x + \frac{2}{3}$.
The slope of this line is $m_2 = -\frac{1}{3}$.
Since the tangent is perpendicular to the line,$m_1 \times m_2 = -1$.
Therefore,$(3x_1^2) \times (-\frac{1}{3}) = -1$.
$-x_1^2 = -1 \implies x_1^2 = 1 \implies x_1 = \pm 1$.
If $x_1 = 1$,then $y_1 = (1)^3 + 5 = 6$.
If $x_1 = -1$,then $y_1 = (-1)^3 + 5 = 4$.
Thus,the points are $(1, 6)$ and $(-1, 4)$.

(B) Given the curve $y^2 = x^3$ at $x = 8$.
Substituting $x = 8$ into the equation: $y^2 = 8^3 = 512$,so $y = \pm \sqrt{512} = \pm 16\sqrt{2}$.
The points are $(8, 16\sqrt{2})$ and $(8, -16\sqrt{2})$.
Differentiating $y^2 = x^3$ with respect to $x$: $2y \frac{dy}{dx} = 3x^2$,which gives $\frac{dy}{dx} = \frac{3x^2}{2y}$.
At $(8, \pm 16\sqrt{2})$,the slope of the tangent $m_t = \frac{3(8)^2}{2(\pm 16\sqrt{2})} = \frac{3 \times 64}{\pm 32\sqrt{2}} = \pm 3\sqrt{2}$.
The slope of the normal $m_n = -\frac{1}{m_t} = \mp \frac{1}{3\sqrt{2}}$.
The equation of the normal is $(y - y_1) = m_n(x - x_1)$.
For $(8, \pm 16\sqrt{2})$: $(y \mp 16\sqrt{2}) = \mp \frac{1}{3\sqrt{2}}(x - 8)$.
Multiplying by $\mp 3\sqrt{2}$: $\mp 3\sqrt{2}y + 96 = x - 8$.
Rearranging gives $x \pm 3\sqrt{2}y = 104$.

(B) Given the curve $y = \sin x$.
First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = \cos x$.
Now,evaluate the slope at the point $(\pi, 0)$: $m = \left( \frac{dy}{dx} \right)_{(\pi, 0)} = \cos(\pi) = -1$.
The equation of the tangent line passing through $(x_1, y_1) = (\pi, 0)$ with slope $m = -1$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - 0 = -1(x - \pi)$.
This simplifies to $y = -x + \pi$,which can be rewritten as $x + y = \pi$.

(A) Differentiating the equation of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$,which simplifies to $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.
$(i)$ If the tangent is parallel to the $x$-axis,then $\frac{dy}{dx} = 0$. This implies $y = 0$. Substituting $y = 0$ into the curve equation,we get $\sqrt{x} = \sqrt{a}$,so $x = a$. The point is $(a, 0)$.
$(ii)$ If the tangent is parallel to the $y$-axis,then $\frac{dy}{dx} = \infty$. This implies $x = 0$. Substituting $x = 0$ into the curve equation,we get $\sqrt{y} = \sqrt{a}$,so $y = a$. The point is $(0, a)$.
$(iii)$ If the tangent makes equal angles with both axes,its slope $\frac{dy}{dx} = \pm 1$. Thus,$-\sqrt{\frac{y}{x}} = \pm 1$,which implies $\frac{y}{x} = 1$,or $y = x$. Substituting $y = x$ into the curve equation,we get $2\sqrt{x} = \sqrt{a}$,so $\sqrt{x} = \frac{\sqrt{a}}{2}$,which gives $x = \frac{a}{4}$. Since $y = x$,$y = \frac{a}{4}$. The point is $(\frac{a}{4}, \frac{a}{4})$.
Thus,the points are $(a, 0), (0, a), (a/4, a/4)$.

(D) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = a(1 + \cos t) = a(2 \cos^2 \frac{t}{2}) = 2a \cos^2 \frac{t}{2}$
$\frac{dy}{dt} = a(\sin t) = 2a \sin \frac{t}{2} \cos \frac{t}{2}$
Now,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \sin \frac{t}{2} \cos \frac{t}{2}}{2a \cos^2 \frac{t}{2}} = \tan \frac{t}{2}$.
The slope of the normal is $-\frac{dx}{dy} = -\cot \frac{t}{2}$.
The length of the normal is given by $|y \sqrt{1 + (\frac{dy}{dx})^2}|$.
$|y \sqrt{1 + \tan^2 \frac{t}{2}}| = |y \sec \frac{t}{2}|$.
Substituting $y = a(1 - \cos t) = 2a \sin^2 \frac{t}{2}$:
Length of normal $= |2a \sin^2 \frac{t}{2} \cdot \frac{1}{\cos \frac{t}{2}}| = |2a \sin \frac{t}{2} \tan \frac{t}{2}|$.

(C) Given the curve $y = 2 \sin x + \sin 2x$.
First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = 2 \cos x + 2 \cos 2x$.
At $x = \pi / 3$,the slope $m$ is:
$m = \left( \frac{dy}{dx} \right)_{x = \pi / 3} = 2 \cos(\pi / 3) + 2 \cos(2\pi / 3) = 2(1/2) + 2(-1/2) = 1 - 1 = 0$.
Next,find the $y$-coordinate at $x = \pi / 3$:
$y = 2 \sin(\pi / 3) + \sin(2\pi / 3) = 2(\sqrt{3}/2) + (\sqrt{3}/2) = \sqrt{3} + \sqrt{3}/2 = 3\sqrt{3}/2$.
The equation of the tangent line at point $(x_1, y_1)$ with slope $m$ is $(y - y_1) = m(x - x_1)$.
Since $m = 0$,the equation becomes $y - 3\sqrt{3}/2 = 0$,which simplifies to $y = 3\sqrt{3}/2$ or $2y = 3\sqrt{3}$.

(A) Given the curve equation $by^2 = (x + a)^3$.
Differentiating with respect to $x$:
$2by \frac{dy}{dx} = 3(x + a)^2$
$\frac{dy}{dx} = \frac{3(x + a)^2}{2by}$
The length of the subnormal $SN$ is given by $|y \frac{dy}{dx}|$:
$SN = y \cdot \frac{3(x + a)^2}{2by} = \frac{3(x + a)^2}{2b}$
The length of the subtangent $ST$ is given by $|y / \frac{dy}{dx}|$:
$ST = y \cdot \frac{2by}{3(x + a)^2} = \frac{2by^2}{3(x + a)^2}$
Since $by^2 = (x + a)^3$,we substitute this into the expression for $ST$:
$ST = \frac{2(x + a)^3}{3(x + a)^2} = \frac{2(x + a)}{3}$
We are given the relation $p(SN) = q(ST)^2$,which implies $\frac{p}{q} = \frac{(ST)^2}{SN}$.
Substituting the expressions for $SN$ and $ST$:
$\frac{p}{q} = \frac{(\frac{2(x + a)}{3})^2}{\frac{3(x + a)^2}{2b}} = \frac{\frac{4(x + a)^2}{9}}{\frac{3(x + a)^2}{2b}} = \frac{4}{9} \cdot \frac{2b}{3} = \frac{8b}{27}$.

(A) Given the curve equation: $y = x^3 - 3x^2 - 9x + 7$.
Differentiating with respect to $x$: $\frac{dy}{dx} = 3x^2 - 6x - 9$.
The tangent is parallel to the $x$-axis when its slope is zero,i.e.,$\frac{dy}{dx} = 0$.
Setting the derivative to zero: $3x^2 - 6x - 9 = 0$.
Dividing by $3$: $x^2 - 2x - 3 = 0$.
Factoring the quadratic equation: $(x - 3)(x + 1) = 0$.
This gives $x = 3$ and $x = -1$.
For $x = -1$: $y = (-1)^3 - 3(-1)^2 - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12$.
For $x = 3$: $y = (3)^3 - 3(3)^2 - 9(3) + 7 = 27 - 27 - 27 + 7 = -20$.
Thus,the points are $(-1, 12)$ and $(3, -20)$.

(A) The given curve is $y = 1 - e^{x/2}$.
To find the point where the curve intersects the $y$-axis,we set $x = 0$.
Substituting $x = 0$ into the equation: $y = 1 - e^{0/2} = 1 - 1 = 0$.
So,the point of intersection is $(0, 0)$.
Now,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{d}{dx}(1 - e^{x/2}) = -e^{x/2} \cdot \frac{1}{2} = -\frac{1}{2}e^{x/2}$.
The slope of the tangent at $(0, 0)$ is $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{2}e^{0} = -\frac{1}{2}$.
The equation of the tangent line passing through $(x_1, y_1) = (0, 0)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$.
$y - 0 = -\frac{1}{2}(x - 0)$.
$y = -\frac{1}{2}x$.
$2y = -x$,which simplifies to $x + 2y = 0$.

(A) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.
At the point $(at^2, 2at)$,the slope of the tangent is $m = \frac{2a}{2at} = \frac{1}{t}$.
The length of the tangent is given by $|y| \sqrt{1 + (\frac{dx}{dy})^2} = |2at| \sqrt{1 + t^2} = 2at\sqrt{t^2 + 1}$ (assuming $t > 0$).
The length of the sub-tangent is given by $|y \frac{dx}{dy}| = |2at \cdot t| = 2at^2$.

(B) The curve is given by $y = be^{-x/a}$.
To find the point where the curve crosses the $y$-axis,we set $x = 0$.
Substituting $x = 0$ into the equation,we get $y = be^0 = b$.
So,the point of intersection is $(0, b)$.
Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = b \cdot (-\frac{1}{a}) e^{-x/a} = -\frac{b}{a} e^{-x/a}$.
At the point $(0, b)$,the slope $m$ is:
$m = (\frac{dy}{dx})_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.
The equation of the tangent line passing through $(0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$:
$y - b = -\frac{b}{a}(x - 0)$.
Multiplying by $a$ and rearranging:
$a(y - b) = -bx \implies ay - ab = -bx \implies bx + ay = ab$.
Dividing both sides by $ab$,we get:
$\frac{bx}{ab} + \frac{ay}{ab} = 1 \implies \frac{x}{a} + \frac{y}{b} = 1$.

(D) Given the curve equation: $y^3 + 3x^2 = 12y$.
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$.
Rearranging to solve for $\frac{dy}{dx}$:
$6x = (12 - 3y^2) \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{6x}{12 - 3y^2} = \frac{2x}{4 - y^2}$.
$A$ tangent is vertical when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator is zero:
$4 - y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$.
Case $1$: If $y = 2$,then $2^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x = \pm \frac{4}{\sqrt{3}}$.
Case $2$: If $y = -2$,then $(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$.
Thus,the points where the tangent is vertical are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

(A) Given the curve $y = \frac{1}{x^2 + 2x + 5}$.
To find the tangent parallel to the $X$-axis,we set the derivative $\frac{dy}{dx} = 0$.
Using the chain rule,$\frac{dy}{dx} = -\frac{1}{(x^2 + 2x + 5)^2} \cdot (2x + 2)$.
Setting $\frac{dy}{dx} = 0$ gives $-(2x + 2) = 0$,which implies $x = -1$.
Now,substitute $x = -1$ into the original equation to find the $y$-coordinate:
$y = \frac{1}{(-1)^2 + 2(-1) + 5} = \frac{1}{1 - 2 + 5} = \frac{1}{4}$.
Since the tangent is parallel to the $X$-axis,its equation is $y = \text{constant}$,which is $y = 1/4$.

(C) Given the curve $xy = 1$,differentiating with respect to $x$ gives $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
The slope of the normal at any point $(x, y)$ on the curve is $m_n = -\frac{1}{dy/dx} = \frac{x}{y}$.
The given 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 $m = -\frac{a}{b}$.
Equating the slopes,we get $-\frac{a}{b} = \frac{x}{y}$,which implies $\frac{a}{b} = -\frac{x}{y}$.
Since $xy = 1$,either both $x, y > 0$ or both $x, y < 0$.
If $x, y > 0$,then $\frac{x}{y} > 0$,so $\frac{a}{b} < 0$. This happens if $a > 0, b < 0$ or $a < 0, b > 0$.
If $x, y < 0$,then $\frac{x}{y} > 0$,so $\frac{a}{b} < 0$. This also happens if $a > 0, b < 0$ or $a < 0, b > 0$.
Thus,the condition is $a < 0, b > 0$ or $a > 0, b < 0$.

(D) Given the curve $y = be^{-x/a}$.
To find the point where it crosses the $y$-axis,set $x = 0$.
Then $y = be^0 = b$.
So,the point of intersection is $(0, b)$.
Now,find the derivative $\frac{dy}{dx} = b \cdot (-\frac{1}{a}) e^{-x/a} = -\frac{b}{a} e^{-x/a}$.
The slope of the tangent at $(0, b)$ is $m = \left(\frac{dy}{dx}\right)_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.
The equation of the tangent line at $(x_1, y_1) = (0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - b = -\frac{b}{a}(x - 0)$.
$y - b = -\frac{b}{a}x$.
Multiply by $a$: $ay - ab = -bx$.
Rearranging: $bx + ay = ab$.
Dividing by $ab$: $\frac{bx}{ab} + \frac{ay}{ab} = \frac{ab}{ab}$.
Thus,$\frac{x}{a} + \frac{y}{b} = 1$.