Tangent and Normal Questions in English

(A) Given the curve $x^4 + y^4 = a^4$.
Let the point of tangency be $P(x_1, y_1)$.
Differentiating with respect to $x$,we get $4x^3 + 4y^3 \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x^3}{y^3}$.
The slope of the tangent at $(x_1, y_1)$ is $m = -\frac{x_1^3}{y_1^3}$.
The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{x_1^3}{y_1^3}(x - x_1)$.
This simplifies to $y y_1^3 - y_1^4 = -x x_1^3 + x_1^4$,or $x x_1^3 + y y_1^3 = x_1^4 + y_1^4$.
Since $(x_1, y_1)$ lies on the curve,$x_1^4 + y_1^4 = a^4$,so the equation is $x x_1^3 + y y_1^3 = a^4$.
The $x$-intercept $p$ is found by setting $y=0$,so $p = \frac{a^4}{x_1^3}$.
The $y$-intercept $q$ is found by setting $x=0$,so $q = \frac{a^4}{y_1^3}$.
Now,calculate $p^{-4/3} + q^{-4/3} = (\frac{a^4}{x_1^3})^{-4/3} + (\frac{a^4}{y_1^3})^{-4/3}$.
$= a^{-16/3} (x_1^4 + y_1^4) = a^{-16/3} (a^4) = a^{4 - 16/3} = a^{-4/3}$.

(D) Let the point on the curve be $(x_1, y_1)$.
Given the curve $ay^2 = x^3$.
Differentiating with respect to $x$,we get $2ay \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{3x^2}{2ay}$.
The slope of the normal at $(x_1, y_1)$ is $m = -\frac{1}{(dy/dx)_{(x_1, y_1)}} = -\frac{2ay_1}{3x_1^2}$.
Since the normal makes equal intercepts on the axes,its slope must be $\pm 1$. Given the geometry of the curve,we set the slope to $-1$ for equal intercepts (i.e.,$x$-intercept = $y$-intercept).
Thus,$-\frac{2ay_1}{3x_1^2} = -1$,which implies $2ay_1 = 3x_1^2$.
From the curve equation,$ay_1^2 = x_1^3$. Squaring the first relation: $4a^2y_1^2 = 9x_1^4$.
Substituting $ay_1^2 = x_1^3$,we get $4a(x_1^3) = 9x_1^4$.
Since $x_1 \neq 0$,we have $4a = 9x_1$,so $x_1 = \frac{4a}{9}$.

(A) Given the curve $y = x(2 - x) = 2x - x^2$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2 - 2x$.
The slope of the tangent at $(2, 0)$ is $m_t = \left( \frac{dy}{dx} \right)_{(2, 0)} = 2 - 2(2) = 2 - 4 = -2$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{-2} = \frac{1}{2}$.
The equation of the normal at $(2, 0)$ is $y - y_1 = m_n(x - x_1)$.
Substituting the values,$y - 0 = \frac{1}{2}(x - 2)$.
$2y = x - 2$,which simplifies to $x - 2y = 2$.

(B) Given the curve $y = a^{1-n}x^n$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = n a^{1-n} x^{n-1}$.
The length of the subnormal at any point $(x, y)$ is given by the formula $L = |y \cdot \frac{dy}{dx}|$.
Substituting the values:
$L = |(a^{1-n}x^n) \cdot (n a^{1-n} x^{n-1})| = |n a^{2(1-n)} x^{2n-1}|$.
For the length of the subnormal to be constant,the exponent of $x$ must be zero.
Therefore,$2n - 1 = 0$,which implies $n = 1/2$.

(C) Given the parametric equations of the curve are $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(1 + \cos \theta) = a(2 \cos^2 \frac{\theta}{2})$
$\frac{dy}{d\theta} = a \sin \theta = 2a \sin \frac{\theta}{2} \cos \frac{\theta}{2}$
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2a \cos^2 \frac{\theta}{2}} = \tan \frac{\theta}{2}$.
Since the tangent makes an angle of $\pi / 4$ with the $x$-axis,the slope is $\tan(\pi / 4) = 1$.
Therefore,$\tan \frac{\theta}{2} = 1$,which implies $\frac{\theta}{2} = \frac{\pi}{4}$,so $\theta = \frac{\pi}{2}$.
Now,substitute $\theta = \frac{\pi}{2}$ into the equations for $x$ and $y$:
$x = a(\frac{\pi}{2} + \sin \frac{\pi}{2}) = a(\frac{\pi}{2} + 1)$
$y = a(1 - \cos \frac{\pi}{2}) = a(1 - 0) = a$
Thus,the required point is $\left( a\left( \frac{\pi}{2} + 1 \right), a \right)$.

(C) Given the curve equation is $y = 2 + \sqrt{4x + 1}$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2 + \sqrt{4x + 1}) = 0 + \frac{1}{2\sqrt{4x + 1}} \times 4 = \frac{2}{\sqrt{4x + 1}}$.
We are given that the slope of the tangent is $\frac{2}{5}$.
So,$\frac{2}{\sqrt{4x + 1}} = \frac{2}{5}$.
This implies $\sqrt{4x + 1} = 5$.
Squaring both sides,we get $4x + 1 = 25$.
$4x = 24$,which gives $x = 6$.
Now,substitute $x = 6$ into the original equation to find $y$:
$y = 2 + \sqrt{4(6) + 1} = 2 + \sqrt{25} = 2 + 5 = 7$.
Thus,the required point is $(6, 7)$.

(A) Given the curve $y = x^3$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = 3x^2$.
According to the problem,the slope of the tangent at a point $(x, y)$ is equal to the $y$-coordinate of that point.
So,$\frac{dy}{dx} = y$.
Substituting the values,we have $3x^2 = x^3$.
This implies $x^3 - 3x^2 = 0$,which factors to $x^2(x - 3) = 0$.
Thus,$x = 0$ or $x = 3$.
If $x = 0$,then $y = 0^3 = 0$. The point is $(0, 0)$.
If $x = 3$,then $y = 3^3 = 27$. The point is $(3, 27)$.
Comparing with the given options,$(3, 27)$ is the correct point.

(D) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.
Differentiating both sides with respect to $x$:
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$.
Solving for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.
For the tangent to be perpendicular to the $x$-axis,the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero.
Thus,$x = 0$.
Substituting $x = 0$ into the original equation $\sqrt{x} + \sqrt{y} = \sqrt{a}$:
$\sqrt{0} + \sqrt{y} = \sqrt{a} \implies \sqrt{y} = \sqrt{a} \implies y = a$.
Therefore,the point is $(0, a)$.

(D) Given the curve $\sqrt{x} + \sqrt{y} = 4$.
Differentiating with respect to $x$,we get:
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
$\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.
At the point $(4, 4)$,the slope $m = -\sqrt{\frac{4}{4}} = -1$.
The equation of the tangent at $(4, 4)$ is $y - 4 = -1(x - 4)$,which simplifies to $y - 4 = -x + 4$,or $x + y = 8$.
To find the intercepts,we write the equation in intercept form: $\frac{x}{8} + \frac{y}{8} = 1$.
The $x$-intercept is $8$ and the $y$-intercept is $8$.
The sum of the intercepts is $8 + 8 = 16$.

(C) The equation of the curve is $y^n = a^{n-1}x$.
Differentiating both sides with respect to $x$,we get:
$n y^{n-1} \frac{dy}{dx} = a^{n-1}$
Therefore,the slope of the tangent at any point $(x_1, y_1)$ is:
$\left( \frac{dy}{dx} \right)_{(x_1, y_1)} = \frac{a^{n-1}}{n y_1^{n-1}}$
The length of the subnormal is given by the formula:
$L = |y_1 \cdot \frac{dy}{dx}| = |y_1 \cdot \frac{a^{n-1}}{n y_1^{n-1}}| = |\frac{a^{n-1}}{n} y_1^{2-n}|$
For the length of the subnormal to be constant,the expression must be independent of $y_1$. This happens when the exponent of $y_1$ is $0$.
Thus,$2 - n = 0$,which implies $n = 2$.

(B) Given the curve $xy = 4$,we can write $y = \frac{4}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.
The equation of the line is $ax + by + c = 0$,which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $m = -\frac{a}{b}$.
Since the line is a tangent to the curve,the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:
$-\frac{4}{x^2} = -\frac{a}{b}$
$\frac{4}{x^2} = \frac{a}{b}$.
Since $x^2 > 0$ for all $x \neq 0$ and $4 > 0$,it follows that $\frac{a}{b} > 0$.
This implies that $a$ and $b$ must have the same sign (both positive or both negative).

(A) Let the point on the curve be $P(x_1, y_1)$. Since $P$ lies on the curve $y = x^2 + 3x$,we have $y_1 = x_1^2 + 3x_1$.
The slope of the tangent at $(x_1, y_1)$ is given by $\frac{dy}{dx} = 2x + 3$.
So,the slope $m = 2x_1 + 3$.
The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = (2x_1 + 3)(x - x_1)$.
Since the tangent passes through $(0, -9)$,we substitute $x = 0$ and $y = -9$:
$-9 - y_1 = (2x_1 + 3)(0 - x_1)$
$-9 - y_1 = -2x_1^2 - 3x_1$
$y_1 = 2x_1^2 + 3x_1 - 9$.
Equating the two expressions for $y_1$:
$x_1^2 + 3x_1 = 2x_1^2 + 3x_1 - 9$
$x_1^2 = 9$
$x_1 = \pm 3$.
If $x_1 = 3$,then $y_1 = (3)^2 + 3(3) = 18$. Point is $(3, 18)$.
If $x_1 = -3$,then $y_1 = (-3)^2 + 3(-3) = 0$. Point is $(-3, 0)$.
Comparing with the given options,the correct point is $(-3, 0)$.

(C) Given $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$.
$\frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta$.
$\frac{dy}{d\theta} = a(\cos \theta - (\cos \theta - \theta \sin \theta)) = a\theta \sin \theta$.
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 $m = -\frac{1}{\tan \theta} = -\cot \theta$.
The equation of the normal at point $\theta$ is $y - y_1 = m(x - x_1)$:
$y - a(\sin \theta - \theta \cos \theta) = -\cot \theta (x - a(\cos \theta + \theta \sin \theta))$.
$y \sin \theta - a \sin^2 \theta + a\theta \cos \theta \sin \theta = -x \cos \theta + a \cos^2 \theta + a\theta \sin \theta \cos \theta$.
$x \cos \theta + y \sin \theta = a(\cos^2 \theta + \sin^2 \theta) = a$.
The distance of this line from the origin $(0,0)$ is $d = \frac{|a|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |a|$,which is constant.

(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 $\frac{dy}{dx}$:
$\frac{dy}{dx} (3y^2 - 12) = -6x \implies \frac{dy}{dx} = -\frac{6x}{3y^2 - 12} = -\frac{2x}{y^2 - 4}$.
For the tangent to be vertical,the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero:
$y^2 - 4 = 0 \implies y^2 = 4 \implies y = \pm 2$.
Case $1$: If $y = 2$,substitute into the original equation:
$(2)^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}}$.
Case $2$: If $y = -2$,substitute into the original equation:
$(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$.
Since $x^2$ cannot be negative for real $x$,there are no real points for $y = -2$.
Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

(B) Given the curve equation: $y = x + \frac{4}{x^{2}}$.
To find the slope of the tangent,we differentiate with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{8}{x^{3}}$.
Since the tangent is parallel to the $x$-axis,its slope must be $0$:
$1 - \frac{8}{x^{3}} = 0$.
Solving for $x$:
$\frac{8}{x^{3}} = 1 \Rightarrow x^{3} = 8 \Rightarrow x = 2$.
Now,find the corresponding $y$-coordinate by substituting $x = 2$ into the original curve equation:
$y = 2 + \frac{4}{2^{2}} = 2 + \frac{4}{4} = 2 + 1 = 3$.
The point of contact is $(2, 3)$.
Since the tangent is parallel to the $x$-axis,its equation is of the form $y = k$. Substituting the $y$-coordinate of the point of contact,we get $y = 3$.

(A) Given the curve $y = \int_{0}^{x} |t| dt$. By the Fundamental Theorem of Calculus,the slope of the tangent is $\frac{dy}{dx} = |x|$.
Since the tangent is parallel to the line $y = 2x$,its slope must be $2$. Thus,$|x| = 2$,which gives $x = 2$ or $x = -2$.
For $x = 2$,$y = \int_{0}^{2} |t| dt = \int_{0}^{2} t dt = [\frac{t^2}{2}]_{0}^{2} = 2$. The point is $(2, 2)$.
The equation of the tangent at $(2, 2)$ is $y - 2 = 2(x - 2) \Rightarrow y = 2x - 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = 2$,so $x = 1$.
For $x = -2$,$y = \int_{0}^{-2} |t| dt = \int_{0}^{-2} (-t) dt = [-\frac{t^2}{2}]_{0}^{-2} = -2$. The point is $(-2, -2)$.
The equation of the tangent at $(-2, -2)$ is $y - (-2) = 2(x - (-2)) \Rightarrow y + 2 = 2x + 4 \Rightarrow y = 2x + 2$. The $x$-intercept is found by setting $y = 0$,giving $2x = -2$,so $x = -1$.
Thus,the $x$-intercepts are $\pm 1$.

(D) Given $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$.
Using $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$,we have:
$f(x) = \tan^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right) = \frac{\pi}{4} + \frac{x}{2}$.
At $x = \frac{\pi}{6}$,$f(\frac{\pi}{6}) = \frac{\pi}{4} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.
The derivative $f'(x) = \frac{1}{2}$.
The slope of the tangent at $x = \frac{\pi}{6}$ is $m_t = \frac{1}{2}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -2$.
The equation of the normal at $(\frac{\pi}{6}, \frac{\pi}{3})$ is $y - \frac{\pi}{3} = -2(x - \frac{\pi}{6})$.
$y - \frac{\pi}{3} = -2x + \frac{\pi}{3} \Rightarrow y = -2x + \frac{2\pi}{3}$.
Checking the points,if $x = 0$,$y = \frac{2\pi}{3}$. Thus,it passes through $(0, \frac{2\pi}{3})$.

(C) The equation of the curve is $y(x - 2)(x - 3) = x + 6$,which can be written as $y = \frac{x + 6}{x^2 - 5x + 6}$.
At the $y$-axis,$x = 0$. Substituting $x = 0$ into the equation,we get $y(0 - 2)(0 - 3) = 0 + 6$,which simplifies to $y(6) = 6$,so $y = 1$. Thus,the point of intersection is $(0, 1)$.
Now,differentiate $y = \frac{x + 6}{x^2 - 5x + 6}$ with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{(x^2 - 5x + 6)(1) - (x + 6)(2x - 5)}{(x^2 - 5x + 6)^2}$.
At the point $(0, 1)$,we have $x = 0$ and $x^2 - 5x + 6 = 6$:
$\frac{dy}{dx} = \frac{(6)(1) - (6)(-5)}{(6)^2} = \frac{6 + 30}{36} = \frac{36}{36} = 1$.
The slope of the tangent at $(0, 1)$ is $1$. Therefore,the slope of the normal is $-\frac{1}{1} = -1$.
The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y - 1 = -x$,or $x + y = 1$.
Checking the options,the point $(\frac{1}{2}, \frac{1}{2})$ satisfies the equation $x + y = 1$ because $\frac{1}{2} + \frac{1}{2} = 1$.

(A) Given the curve equation: ${y^n} = {a^{n - 1}}x$.
Differentiating both sides with respect to $x$,we get: $n{y^{n - 1}}\frac{{dy}}{{dx}} = {a^{n - 1}}$.
Thus,the slope of the tangent is: $\frac{{dy}}{{dx}} = \frac{{{a^{n - 1}}}}{{n{y^{n - 1}}}}$.
The length of the subnormal is given by the formula: $\text{Subnormal} = y \left| \frac{{dy}}{{dx}} \right|$.
Substituting the value of $\frac{{dy}}{{dx}}$: $\text{Subnormal} = y \cdot \frac{{{a^{n - 1}}}}{{n{y^{n - 1}}}} = \frac{{{a^{n - 1}}}}{n} \cdot y^{1 - (n - 1)} = \frac{{{a^{n - 1}}}}{n} \cdot y^{2 - n}$.
For the subnormal to be constant,it must be independent of $y$.
This implies the exponent of $y$ must be $0$,so $2 - n = 0$,which gives $n = 2$.

(C) The equation of the curve is $x^2 y^2 = a^4$.
Differentiating both sides with respect to $x$,we get:
$x^2 \cdot 2y \frac{dy}{dx} + y^2 \cdot 2x = 0$
$\frac{dy}{dx} = -\frac{2xy^2}{2x^2y} = -\frac{y}{x}$
At the point $(-a, a)$,the slope of the tangent is:
$\left( \frac{dy}{dx} \right)_{(-a, a)} = -\frac{a}{-a} = 1$
The length of the subtangent is given by the formula $\left| \frac{y}{dy/dx} \right|$.
Substituting the values,we get:
$\text{Subtangent} = \left| \frac{a}{1} \right| = a$.

(C) Given the curve $f(x) = x^2 + bx - b$. Since the point $(1, 1)$ lies on the curve,we have $1 = 1^2 + b(1) - b$,which is $1 = 1$,confirming the point is on the curve for any $b$.
First,find the derivative: $\frac{dy}{dx} = 2x + b$.
At the point $(1, 1)$,the slope of the tangent is $m = 2(1) + b = 2 + b$.
The equation of the tangent line at $(1, 1)$ is $y - 1 = (2 + b)(x - 1)$.
Simplifying,$y - 1 = (2 + b)x - (2 + b)$,which gives $(2 + b)x - y = 1 + b$.
To find the intercepts,set $y = 0$ to get $x = \frac{1 + b}{2 + b}$ (point $A$),and set $x = 0$ to get $y = -(1 + b)$ (point $B$).
Since the triangle lies in the first quadrant,the intercepts must be positive: $\frac{1 + b}{2 + b} > 0$ and $-(1 + b) > 0$.
This implies $1 + b < 0$ and $2 + b < 0$,so $b < -2$.
The area of the triangle is $\frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times \left| \frac{1 + b}{2 + b} \right| \times |-(1 + b)| = 2$.
Since $b < -2$,$1 + b$ is negative and $2 + b$ is negative,so $\frac{1 + b}{2 + b} > 0$ and $-(1 + b) > 0$.
Thus,$\frac{1}{2} \cdot \frac{1 + b}{2 + b} \cdot (-(1 + b)) = 2$.
$-(1 + b)^2 = 4(2 + b) \Rightarrow -(1 + 2b + b^2) = 8 + 4b \Rightarrow b^2 + 6b + 9 = 0$.
$(b + 3)^2 = 0$,so $b = -3$. This satisfies $b < -2$.

(A) The curve is $y = \sqrt{x}$,which implies $y^2 = x$ for $y \ge 0$.
The slope of the tangent to the curve at any point $(x, y)$ is $\frac{dy}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2y}$.
The slope of the normal at a point $(x_0, y_0)$ on the curve is $m = -\frac{1}{(dy/dx)} = -2y_0$.
Since the normal passes through $(3, 6)$,the slope is $m = \frac{y_0 - 6}{x_0 - 3}$.
Equating the slopes: $-2y_0 = \frac{y_0 - 6}{y_0^2 - 3}$ (since $x_0 = y_0^2$).
$-2y_0(y_0^2 - 3) = y_0 - 6$ $\Rightarrow -2y_0^3 + 6y_0 = y_0 - 6$ $\Rightarrow 2y_0^3 - 5y_0 - 6 = 0$.
Testing for roots,$y_0 = 2$ satisfies the equation: $2(8) - 5(2) - 6 = 16 - 10 - 6 = 0$.
For $y_0 = 2$,the point on the curve is $(4, 2)$ and the slope of the normal is $m = -2(2) = -4$.
The equation of the line is $y - 6 = -4(x - 3)$ $\Rightarrow y - 6 = -4x + 12$ $\Rightarrow 4x + y - 18 = 0$.

(B) Given the equation of the circle: $x^2 + y^2 = R^2$.
Differentiating with respect to $x$: $2x + 2yy' = 0$,which simplifies to $x + yy' = 0$,so $y' = -\frac{x}{y}$.
Differentiating again with respect to $x$: $1 + y'y' + yy'' = 0$,which gives $1 + (y')^2 + yy'' = 0$.
Therefore,$y'' = -\frac{1 + (y')^2}{y}$.
The expression for $k$ is given by $k = \frac{y''}{(1 + (y')^2)^{3/2}}$.
Substituting $y''$: $k = \frac{-(1 + (y')^2)}{y(1 + (y')^2)^{3/2}} = -\frac{1}{y\sqrt{1 + (y')^2}}$.
Since $y' = -\frac{x}{y}$,we have $1 + (y')^2 = 1 + \frac{x^2}{y^2} = \frac{y^2 + x^2}{y^2} = \frac{R^2}{y^2}$.
Thus,$\sqrt{1 + (y')^2} = \sqrt{\frac{R^2}{y^2}} = \frac{R}{|y|}$.
Substituting this back into $k$: $k = -\frac{1}{y \cdot (R/|y|)} = -\frac{1}{R} \cdot \frac{|y|}{y}$.
Assuming the upper semicircle where $y > 0$,we get $k = -\frac{1}{R}$.

(B) The curve is given by $y = K e^{Kx}$.
To find the slope of the tangent at the point of intersection with the $y$-axis,we set $x = 0$.
At $x = 0$,$y = K e^{K(0)} = K$.
The derivative is $\frac{dy}{dx} = K \cdot K e^{Kx} = K^2 e^{Kx}$.
At $x = 0$,the slope $m = \left. \frac{dy}{dx} \right|_{x=0} = K^2 e^0 = K^2$.
Let $\psi$ be the angle the tangent makes with the positive $x$-axis. Then $\tan \psi = m = K^2$.
The angle $\theta$ that the curve makes with the $y$-axis is the complement of the angle $\psi$ that the tangent makes with the $x$-axis,so $\theta = \frac{\pi}{2} - \psi$.
Thus,$\tan \theta = \tan(\frac{\pi}{2} - \psi) = \cot \psi = \frac{1}{\tan \psi} = \frac{1}{K^2}$.
Therefore,$\theta = \tan^{-1}(\frac{1}{K^2}) = \cot^{-1}(K^2)$.
Comparing this with the given options,the correct option is $B$.

(D) Given $f(x) = \int_{2}^{x} (2t - 5) \, dt$.
Evaluating the integral: $f(x) = [t^2 - 5t]_{2}^{x} = (x^2 - 5x) - (4 - 10) = x^2 - 5x + 6$.
The graph cuts the $x$-axis where $f(x) = 0$.
$x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0$,so $x = 2$ and $x = 3$.
The points of intersection are $(2, 0)$ and $(3, 0)$.
The slope of the tangent is $f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$.
At $x = 2$,$m_1 = 2(2) - 5 = -1$.
At $x = 3$,$m_2 = 2(3) - 5 = 1$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{0} \right| = \infty$.
Thus,$\theta = \frac{\pi}{2}$.

(C) Given the curve equation $x^2y = c^3$.
Differentiating with respect to $x$,we get $x^2 \frac{dy}{dx} + 2xy = 0$,which implies $\frac{dy}{dx} = -\frac{2y}{x}$.
The equation of the tangent at point $(x, y)$ is $Y - y = -\frac{2y}{x}(X - x)$.
To find the $x$-intercept $a$,set $Y = 0$: $-y = -\frac{2y}{x}(a - x) \implies x = 2(a - x) \implies 3x = 2a \implies a = \frac{3x}{2}$.
To find the $y$-intercept $b$,set $X = 0$: $b - y = -\frac{2y}{x}(0 - x) \implies b - y = 2y \implies b = 3y$.
Now,calculate $a^2b$: $a^2b = (\frac{3x}{2})^2(3y) = \frac{9x^2}{4} \cdot 3y = \frac{27}{4}x^2y$.
Since $x^2y = c^3$,we have $a^2b = \frac{27}{4}c^3$.

(C) The equation of the curve is $xy^n = a^{n+1}$.
Taking the derivative with respect to $x$:
$y^n + x \cdot n y^{n-1} \frac{dy}{dx} = 0$
$x n y^{n-1} \frac{dy}{dx} = -y^n$
$\frac{dy}{dx} = -\frac{y^n}{n x y^{n-1}} = -\frac{y}{nx}$.
The length of the subnormal is given by $|y \frac{dy}{dx}|$.
Substituting $\frac{dy}{dx}$:
$|y \cdot (-\frac{y}{nx})| = |-\frac{y^2}{nx}|$.
From the original equation,$y^n = \frac{a^{n+1}}{x}$,so $y^2 = (\frac{a^{n+1}}{x})^{2/n} = \frac{a^{2(n+1)/n}}{x^{2/n}}$.
Substituting this into the subnormal expression:
$|\frac{y^2}{nx}| = |\frac{a^{2(n+1)/n}}{n x \cdot x^{2/n}}| = |\frac{a^{2(n+1)/n}}{n x^{(n+2)/n}}|$.
For the subnormal to be constant,the power of $x$ must be $0$,so $\frac{n+2}{n} = 0$,which implies $n = -2$.
Thus,for $n = -2$,the subnormal is constant.

(B) Let the tangent at point $P(x_1, y_1)$ intersect the $X$-axis at $A(a, 0)$ and the $Y$-axis at $B(0, -2a)$ (since the segment on $OX$ is half that on $OY$,and $B$ is on the negative $Y$-axis).
The slope of the tangent is $m = \frac{0 - (-2a)}{a - 0} = \frac{2a}{a} = 2$.
Given $y = x^3 - 3x^2 - 7x + 6$,the slope at any point is $\frac{dy}{dx} = 3x^2 - 6x - 7$.
Equating the slope to $2$: $3x_1^2 - 6x_1 - 7 = 2 \implies 3x_1^2 - 6x_1 - 9 = 0 \implies x_1^2 - 2x_1 - 3 = 0$.
Solving for $x_1$: $(x_1 - 3)(x_1 + 1) = 0$,so $x_1 = 3$ or $x_1 = -1$.
If $x_1 = 3$,$y_1 = (3)^3 - 3(3)^2 - 7(3) + 6 = 27 - 27 - 21 + 6 = -15$. Point is $(3, -15)$.
If $x_1 = -1$,$y_1 = (-1)^3 - 3(-1)^2 - 7(-1) + 6 = -1 - 3 + 7 + 6 = 9$. Point is $(-1, 9)$.
Checking the intercept condition for $P(-1, 9)$: The tangent equation is $y - 9 = 2(x + 1) \implies y = 2x + 11$. $X$-intercept $A$ is $(-5.5, 0)$,$Y$-intercept $B$ is $(0, 11)$. The length on $OX$ is $5.5$ (negative side),which does not satisfy the condition.
Checking $P(3, -15)$: The tangent equation is $y + 15 = 2(x - 3) \implies y = 2x - 21$. $X$-intercept $A$ is $(10.5, 0)$,$Y$-intercept $B$ is $(0, -21)$. The length on $OX$ is $10.5$,and on $OY$ is $21$. Since $10.5 = \frac{1}{2}(21)$,this point satisfies the condition.
Thus,the correct point is $(3, -15)$.

(C) Given the curve equation: $\frac{a}{x^2} + \frac{b}{y^2} = 1$.
Differentiating both sides with respect to $x$:
$-\frac{2a}{x^3} - \frac{2b}{y^3} \frac{dy}{dx} = 0$.
This gives the slope of the tangent: $\frac{dy}{dx} = -\frac{ay^3}{bx^3}$.
The equation of the tangent at point $(x, y)$ is: $Y - y = -\frac{ay^3}{bx^3}(X - x)$.
To find the $x-$intercept,set $Y = 0$:
$-y = -\frac{ay^3}{bx^3}(X - x) \implies X - x = \frac{bx^3}{ay^2} \implies X = x + \frac{bx^3}{ay^2}$.
Simplifying the expression: $X = x \left( 1 + \frac{bx^2}{ay^2} \right) = x \left( \frac{ay^2 + bx^2}{ay^2} \right)$.
Since $\frac{a}{x^2} + \frac{b}{y^2} = 1$,we have $ay^2 + bx^2 = x^2y^2$.
Substituting this into the expression for $X$: $X = x \left( \frac{x^2y^2}{ay^2} \right) = \frac{x^3}{a}$.
Thus,the $x-$intercept is proportional to the cube of the abscissa $x^3$.

(D) To find the slope of the tangent at the origin $(0,0)$,we consider the lowest degree terms of the curves.
For the first curve $y^3 - x^2y + 5y - 2x = 0$,the lowest degree terms are $5y - 2x = 0$.
Thus,$5y = 2x$,which gives $y = \frac{2}{5}x$. The slope $m_1 = \frac{2}{5}$.
For the second curve $x^4 - x^3y^2 + 5x + 2y = 0$,the lowest degree terms are $5x + 2y = 0$.
Thus,$2y = -5x$,which gives $y = -\frac{5}{2}x$. The slope $m_2 = -\frac{5}{2}$.
Since $m_1 \cdot m_2 = (\frac{2}{5}) \cdot (-\frac{5}{2}) = -1$,the tangents are perpendicular to each other.
Therefore,the angle $\theta$ between the tangents is $\frac{\pi}{2}$.

(A) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.
At a point $(x_1, y_1)$,the slope of the tangent is $m_T = \frac{x_1^2}{6y_1}$.
The slope of the normal is $m_N = -\frac{1}{m_T} = -\frac{6y_1}{x_1^2}$.
Since the normal makes equal intercepts with the axes,its slope must be $\pm 1$. Given the geometry,the slope of the normal is $-1$.
So,$-\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.
Substitute this into the curve equation $9y_1^2 = x_1^3$. Since $x_1^2 = 6y_1$,we have $x_1^2 = 6(\frac{x_1^3}{9}) = \frac{2}{3}x_1^3$.
For $x_1 \neq 0$,$1 = \frac{2}{3}x_1 \implies x_1 = \frac{3}{2}$. However,checking the slope condition again,if the normal makes equal intercepts,its equation is $x + y = c$ or $x - y = c$. The slope is $-1$.
Solving $x_1^2 = 6y_1$ and $9y_1^2 = x_1^3$ simultaneously: $9(\frac{x_1^2}{6})^2 = x_1^3 \implies 9(\frac{x_1^4}{36}) = x_1^3 \implies \frac{x_1^4}{4} = x_1^3 \implies x_1 = 4$.
Then $y_1^2 = \frac{4^3}{9} = \frac{64}{9} \implies y_1 = \pm \frac{8}{3}$.
Thus,the points are $(4, \frac{8}{3})$ and $(4, -\frac{8}{3})$. Option $A$ matches.

(C) Given the curve $y = x^n$.
The slope of the tangent at point $P(a, a^n)$ is $\frac{dy}{dx} = n x^{n-1} = n a^{n-1}$.
The slope of the normal at point $P$ is $m_n = -\frac{1}{n a^{n-1}}$.
The equation of the normal line passing through $(a, a^n)$ is $y - a^n = -\frac{1}{n a^{n-1}}(x - a)$.
To find the $y-$ intercept $b$,we set $x = 0$ in the equation:
$b - a^n = -\frac{1}{n a^{n-1}}(0 - a)$
$b - a^n = \frac{a}{n a^{n-1}} = \frac{1}{n a^{n-2}}$
$b = a^n + \frac{1}{n a^{n-2}}$.
Now,we evaluate the limit $\mathop {Lim}\limits_{a \to 0} b = \mathop {Lim}\limits_{a \to 0} (a^n + \frac{1}{n a^{n-2}})$.
If $n=2$,then $b = a^2 + \frac{1}{2 a^0} = a^2 + \frac{1}{2}$.
As $a \to 0$,$b \to 0 + \frac{1}{2} = \frac{1}{2}$.
Thus,$n = 2$ satisfies the given condition.

(B) Let the tangent line be $y = mx + c$.
For $x \ge 0$,$f(x) = x^2 + 8$. Setting $mx + c = x^2 + 8$,we get $x^2 - mx + (8 - c) = 0$. For tangency,the discriminant $D = 0$,so $m^2 - 4(8 - c) = 0$,which implies $m^2 = 32 - 4c$ (Equation $1$).
For $x < 0$,$f(x) = -x^2$. Setting $mx + c = -x^2$,we get $x^2 + mx + c = 0$. For tangency,$D = 0$,so $m^2 - 4c = 0$,which implies $m^2 = 4c$ (Equation $2$).
Equating the two expressions for $m^2$: $4c = 32 - 4c$,which gives $8c = 32$,so $c = 4$.
Substituting $c = 4$ into Equation $2$,$m^2 = 16$,so $m = \pm 4$. Since the tangent must touch both branches,we consider the line $y = 4x + 4$.
To find the $x$-intercept,set $y = 0$: $0 = 4x + 4$,which gives $x = -1$.

(A) Given the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$.
At the point where the abscissa $x = a$,we have $(\frac{a}{a})^n + (\frac{y}{b})^n = 2$,which implies $1 + (\frac{y}{b})^n = 2$,so $(\frac{y}{b})^n = 1$.
Since $n \in N$,this gives $y = b$ (assuming $n$ is odd or considering the principal value).
Thus,the point of tangency is $(a, b)$.
Differentiating the curve with respect to $x$:
$n(\frac{x}{a})^{n-1} \cdot \frac{1}{a} + n(\frac{y}{b})^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$.
At $(a, b)$,this becomes $n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$.
$\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{b}{a}$.
The equation of the tangent at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$.
$a(y - b) = -b(x - a) \implies ay - ab = -bx + ab$.
$bx + ay = 2ab \implies \frac{x}{a} + \frac{y}{b} = 2$.

(D) Given the curve equation: $\sqrt{xy} = a + x$. Squaring both sides,we get $xy = (a + x)^2 = a^2 + x^2 + 2ax$.
Solving for $y$,we have $y = \frac{a^2}{x} + x + 2a$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2} + 1$.
Since the tangent cuts off equal intercepts from the coordinate axes,its slope must be $\pm 1$. However,for a line to cut off equal intercepts $c$ and $c$ (where $c \neq 0$),the slope is $-1$.
Setting the derivative equal to $-1$: $-\frac{a^2}{x^2} + 1 = -1$.
This simplifies to $\frac{a^2}{x^2} = 2$,which implies $x^2 = \frac{a^2}{2}$.
Thus,$x = \pm \frac{a}{\sqrt{2}}$.
Since $a > 0$,both values are valid abscissae.

(D) Given the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$.
The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x^2 - 5x + 7$.
Since the intercepts are equal in magnitude but opposite in sign,the slope of the tangent must be $-1$ or $1$.
Case $1$: Slope $m = -1$.
$x^2 - 5x + 7 = -1 \implies x^2 - 5x + 8 = 0$. The discriminant $D = 25 - 32 = -7 < 0$,so no real solution.
Case $2$: Slope $m = 1$.
$x^2 - 5x + 7 = 1 \implies x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0$.
Thus,$x = 2$ or $x = 3$.
For $x = 2$,$f(2) = \frac{8}{3} - \frac{5(4)}{2} + 7(2) - 4 = \frac{8}{3} - 10 + 14 - 4 = \frac{8}{3}$. Point is $(2, 8/3)$.
For $x = 3$,$f(3) = \frac{27}{3} - \frac{5(9)}{2} + 7(3) - 4 = 9 - 22.5 + 21 - 4 = 3.5 = 7/2$. Point is $(3, 7/2)$.
Both points satisfy the condition. Therefore,the correct option is $(D)$.

(A) Let the point $P$ be $(x, y)$. The equation of the tangent at $P$ is $Y - y = f'(x)(X - x)$.
For point $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$,so $A = (x - \frac{y}{f'(x)}, 0)$.
For point $B$ (where $X=0$),$Y = y - x f'(x)$,so $B = (0, y - x f'(x))$.
The normal at $P$ has slope $-\frac{1}{f'(x)}$. Its equation is $Y - y = -\frac{1}{f'(x)}(X - x)$.
For point $C$ (where $X=0$),$Y = y + \frac{x}{f'(x)}$,so $C = (0, y + \frac{x}{f'(x)})$.
Given $AC = BC$,which implies $AC^2 = BC^2$. Since $A$ is on the $x$-axis and $B, C$ are on the $y$-axis,the distance formula gives:
$(x - \frac{y}{f'(x)})^2 + (y + \frac{x}{f'(x)})^2 = (y - x f'(x) - (y + \frac{x}{f'(x)}))^2$
$(x - \frac{y}{f'(x)})^2 + (y + \frac{x}{f'(x)})^2 = (-x f'(x) - \frac{x}{f'(x)})^2$
$x^2 - \frac{2xy}{f'(x)} + \frac{y^2}{(f'(x))^2} + y^2 + \frac{2xy}{f'(x)} + \frac{x^2}{(f'(x))^2} = x^2 (f'(x) + \frac{1}{f'(x)})^2$
$x^2 + y^2 + \frac{x^2 + y^2}{(f'(x))^2} = x^2 (f'(x)^2 + 2 + \frac{1}{f'(x)^2})$
$(x^2 + y^2)(1 + \frac{1}{(f'(x))^2}) = x^2 (f'(x)^2 + 2 + \frac{1}{f'(x)^2})$
This simplifies to $f'(x) = -\frac{y}{x}$.
Integrating $\frac{dy}{y} = -\frac{dx}{x}$ gives $\ln y = -\ln x + \ln c$,so $xy = c$.
Using $f(2) = 3$,we get $2 \times 3 = c$,so $c = 6$.
The equation is $xy = 6$ or $y = \frac{6}{x}$.

(B) Let the point of tangency be $P(x_1, y_1)$ on the curve $xy = 100$.
Since $P$ lies on the curve,$y_1 = \frac{100}{x_1}$.
Differentiating $xy = 100$ with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
At point $P(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1} = -\frac{100/x_1}{x_1} = -\frac{100}{x_1^2}$.
The equation of the tangent at $P(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$.
$y - y_1 = -\frac{y_1}{x_1}x + y_1 \implies \frac{y_1}{x_1}x + y = 2y_1$.
Dividing by $2y_1$,we get $\frac{x}{2x_1} + \frac{y}{2y_1} = 1$.
The $x$-intercept is $A(2x_1, 0)$ and the $y$-intercept is $B(0, 2y_1)$.
The area of $\Delta OAB = \frac{1}{2} \times |2x_1| \times |2y_1| = 2|x_1 y_1|$.
Since $x_1 y_1 = 100$,the area is $2 \times 100 = 200$ square units.

(A) For the curve $y = x^2 + 6x + 10$,the derivative is $\frac{dy}{dx} = 2x + 6$.
At point $(-2, 2)$,the slope of the tangent is $m_t = 2(-2) + 6 = 2$.
Therefore,the slope of the normal at $(-2, 2)$ is $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
For the curve $y = ax^2 + bx + \frac{7}{2}$,the point $(1, 2)$ lies on the curve,so $2 = a(1)^2 + b(1) + \frac{7}{2}$,which simplifies to $a + b = 2 - \frac{7}{2} = -\frac{3}{2}$ (Equation $1$).
The derivative is $\frac{dy}{dx} = 2ax + b$. At $x = 1$,the slope of the tangent is $2a + b$.
Since this tangent is parallel to the normal of the first curve,$2a + b = -\frac{1}{2}$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(2a + b) - (a + b) = -\frac{1}{2} - (-\frac{3}{2}) \Rightarrow a = 1$.
Substituting $a = 1$ into Equation $1$: $1 + b = -\frac{3}{2} \Rightarrow b = -\frac{3}{2} - 1 = -\frac{5}{2}$.

(A) Let $P(x_1, y_1)$ be a point on the curve $y = f(x)$.
Let $m = \frac{dy}{dx}$ be the slope of the tangent at $P$.
The length of the subtangent is given by $\left| \frac{y_1}{m} \right|$ and the length of the subnormal is given by $|y_1 m|$.
Given that the subtangent and subnormal are equal,we have:
$\left| \frac{y_1}{m} \right| = |y_1 m|$
$\Rightarrow \frac{1}{|m|} = |m|$
$\Rightarrow m^2 = 1 \Rightarrow m = \pm 1$.
The length of the normal at point $P$ is given by $|y_1| \sqrt{1 + m^2}$.
Substituting $m^2 = 1$,we get:
$\text{Length of normal} = |y_1| \sqrt{1 + 1} = |y_1| \sqrt{2} = \sqrt{2} \times \text{ordinate}$.

(C) Given the curve $x^{2/3} + y^{2/3} = a^{2/3}$.
Let the point of tangency be $(x_1, y_1)$.
Differentiating with respect to $x$:
$\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\left(\frac{y_1}{x_1}\right)^{1/3}$.
The equation of the tangent at $(x_1, y_1)$ is:
$y - y_1 = -\left(\frac{y_1}{x_1}\right)^{1/3}(x - x_1)$
$\frac{y}{y_1^{1/3}} + \frac{x}{x_1^{1/3}} = \frac{y_1}{y_1^{1/3}} + \frac{x_1}{x_1^{1/3}} = y_1^{2/3} + x_1^{2/3} = a^{2/3}$.
Dividing by $a^{2/3}$:
$\frac{x}{a^{2/3}x_1^{1/3}} + \frac{y}{a^{2/3}y_1^{1/3}} = 1$.
The $x$-intercept is $OA = a^{2/3}x_1^{1/3}$ and the $y$-intercept is $OB = a^{2/3}y_1^{1/3}$.
The sum of the squares of the intercepts is:
$OA^2 + OB^2 = (a^{2/3}x_1^{1/3})^2 + (a^{2/3}y_1^{1/3})^2$
$= a^{4/3}x_1^{2/3} + a^{4/3}y_1^{2/3}$
$= a^{4/3}(x_1^{2/3} + y_1^{2/3})$
$= a^{4/3}(a^{2/3}) = a^2$.

(C) Given the curve equation: $y^n = a^{n-1}x$.
Differentiating both sides with respect to $x$:
$n y^{n-1} \frac{dy}{dx} = a^{n-1}$.
Thus,$\frac{dy}{dx} = \frac{a^{n-1}}{n y^{n-1}}$.
The length of the subnormal is given by the formula: $L = |y \frac{dy}{dx}|$.
Substituting the value of $\frac{dy}{dx}$:
$L = |y \cdot \frac{a^{n-1}}{n y^{n-1}}| = |\frac{a^{n-1}}{n} \cdot y^{1-(n-1)}| = |\frac{a^{n-1}}{n} \cdot y^{2-n}|$.
For the length of the subnormal to be constant,the expression must be independent of $y$.
This occurs when the exponent of $y$ is zero: $2 - n = 0$,which implies $n = 2$.

(D) Let $P = (t_1, t_1^3 - t_1)$ and $Q = (t_2, t_2^3 - t_2)$.
The slope of the tangent at $P$ is given by $\frac{dy}{dx} = 3x^2 - 1$,so at $P$,$m = 3t_1^2 - 1$.
The slope of the chord $PQ$ is $\frac{(t_2^3 - t_2) - (t_1^3 - t_1)}{t_2 - t_1} = \frac{(t_2 - t_1)(t_2^2 + t_1t_2 + t_1^2) - (t_2 - t_1)}{t_2 - t_1} = t_2^2 + t_1t_2 + t_1^2 - 1$.
Since the tangent at $P$ intersects the curve at $Q$,the slope of the tangent equals the slope of the chord $PQ$:
$3t_1^2 - 1 = t_2^2 + t_1t_2 + t_1^2 - 1$
$2t_1^2 - t_1t_2 - t_2^2 = 0$
$(2t_1 + t_2)(t_1 - t_2) = 0$.
Since $P$ and $Q$ are distinct,$t_1 \neq t_2$,so $t_2 = -2t_1$.
The slope $m_{OP} = \frac{t_1^3 - t_1}{t_1} = t_1^2 - 1$ and $m_{OQ} = \frac{t_2^3 - t_2}{t_2} = t_2^2 - 1$.
We need to calculate $\frac{m_{OQ} + 1}{m_{OP} + 1} = \frac{(t_2^2 - 1) + 1}{(t_1^2 - 1) + 1} = \frac{t_2^2}{t_1^2}$.
Substituting $t_2 = -2t_1$,we get $\frac{(-2t_1)^2}{t_1^2} = \frac{4t_1^2}{t_1^2} = 4$.

(B) Given curves are $C_1: \frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $C_2: y^3 = 16x$.
For $C_2$,differentiating with respect to $x$: $3y^2 \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = m_1 = \frac{16}{3y^2}$.
For $C_1$,differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = -\frac{4x}{a^2y}$.
Since the curves intersect orthogonally,$m_1 \times m_2 = -1$.
$\left(\frac{16}{3y^2}\right) \times \left(-\frac{4x}{a^2y}\right) = -1$.
$\frac{64x}{3a^2y^3} = 1 \implies 64x = 3a^2y^3$.
Substitute $y^3 = 16x$ into the equation: $64x = 3a^2(16x)$.
Assuming $x \neq 0$,we get $64 = 48a^2$.
$a^2 = \frac{64}{48} = \frac{4}{3}$.

(B) Given the curve $y = e^{2x} + x^2$.
At $x = 0$,$y = e^0 + 0^2 = 1 + 0 = 1$.
So,the point of contact is $(0, 1)$.
Now,find the derivative $\frac{dy}{dx} = 2e^{2x} + 2x$.
At $x = 0$,the slope of the tangent $m_t = 2e^0 + 2(0) = 2$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
The equation of the normal at $(0, 1)$ is $y - 1 = -\frac{1}{2}(x - 0)$,which simplifies to $y = -\frac{1}{2}x + 1$ or $x + 2y = 2$.
The normal intersects the $x$-axis at $y = 0$,so $x + 2(0) = 2 \implies x = 2$. The point is $(2, 0)$.
The normal intersects the $y$-axis at $x = 0$,so $0 + 2y = 2 \implies y = 1$. The point is $(0, 1)$.
The area of the triangle formed by the axes and the normal is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$.

(C) Given the curve $x^{2/3} + y^{2/3} = 2$.
Differentiating with respect to $x$,we get $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0$.
At the point $(1, 1)$,the slope of the tangent is $\frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\frac{1^{-1/3}}{1^{-1/3}} = -1$.
The equation of the tangent at $(1, 1)$ is $y - 1 = -1(x - 1)$,which simplifies to $x + y = 2$.
The intercept form of the line is $\frac{x}{2} + \frac{y}{2} = 1$.
The $x$-intercept is $2$ and the $y$-intercept is $2$.
The sum of the intercepts is $2 + 2 = 4$.

(B) The given curve is $y = \frac{1}{x}$.
$\frac{dy}{dx} = -\frac{1}{x^{2}}$.
The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^{2}}$.
The slope of the normal is the negative reciprocal of the tangent slope,which is $x^{2}$.
Since $x^{2} > 0$ for all $x \neq 0$,the slope of the normal must be positive.
The given line is $(\log_{2}(1 + 5a - a^{2}))x - 5y - (a^{2} - 5) = 0$,which can be written as $y = \frac{\log_{2}(1 + 5a - a^{2})}{5}x - \frac{a^{2} - 5}{5}$.
Thus,the slope of the line is $m = \frac{\log_{2}(1 + 5a - a^{2})}{5}$.
Since $m > 0$,we have $\log_{2}(1 + 5a - a^{2}) > 0$.
This implies $1 + 5a - a^{2} > 2^{0} = 1$.
$5a - a^{2} > 0 \Rightarrow a^{2} - 5a < 0$.
$a(a - 5) < 0$,which gives $a \in (0, 5)$.

(B) Given the curve $xy = 4$,we can write $y = \frac{4}{x}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.
At the point $(2,2)$,the slope of the tangent is $m_t = -\frac{4}{2^2} = -1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal at $(2,2)$ is $y - 2 = 1(x - 2)$,which simplifies to $y = x$.
To find the point where the normal meets the curve again,we substitute $y = x$ into the equation of the curve $xy = 4$:
$x(x) = 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Since $x = 2$ is the original point,the other point is $x = -2$.
Substituting $x = -2$ into $y = x$,we get $y = -2$.
Thus,the normal meets the curve again at the point $(-2, -2)$.

(C) The slope of the normal $m_n$ is given by $\tan(\theta)$,where $\theta = \frac{2\pi}{3}$.
$m_n = \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$.
We know that the slope of the normal is related to the derivative $f'(x)$ by the formula $m_n = -\frac{1}{f'(x)}$.
Substituting the values at the point $(4, 6)$:
$-\sqrt{3} = -\frac{1}{f'(4)}$.
Solving for $f'(4)$:
$f'(4) = \frac{1}{\sqrt{3}}$.