Tangent and Normal Questions in English

(A) Given $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$.
First,find the derivative $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \cos^2(\theta/2)} = \tan(\theta/2)$.
Length of the tangent is given by $|y \sqrt{1 + (dx/dy)^2}|$ or $|y \csc(\theta/2)|$. Specifically,the length of the tangent is $|y \frac{\sqrt{1 + (dy/dx)^2}}{dy/dx}| = |\frac{y \sec(\theta/2)}{\tan(\theta/2)}| = |\frac{a(1 - \cos \theta)}{\sin(\theta/2)}| = |\frac{2a \sin^2(\theta/2)}{\sin(\theta/2)}| = 2a \sin(\theta/2)$.
Length of the sub-tangent is given by $|y / (dy/dx)| = |\frac{a(1 - \cos \theta)}{\tan(\theta/2)}| = |\frac{2a \sin^2(\theta/2)}{\sin(\theta/2) / \cos(\theta/2)}| = |2a \sin(\theta/2) \cos(\theta/2)| = a \sin \theta$.

(A) Let the point on the curve $9y^2 = x^3$ be $P(x_1, y_1)$.
Differentiating with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.
The slope of the tangent at $P(x_1, y_1)$ is $m_t = \frac{x_1^2}{6y_1}$.
The slope of the normal at $P(x_1, y_1)$ 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 $\pm 1$.
Case $1$: $m_n = -1 \implies -\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.
Since $P$ lies on the curve,$9y_1^2 = x_1^3$. Substituting $x_1^2 = 6y_1$,we get $9y_1^2 = x_1(6y_1) \implies 9y_1^2 = 6x_1y_1$. Assuming $y_1 \neq 0$,$9y_1 = 6x_1 \implies y_1 = \frac{2}{3}x_1$.
Substituting into $x_1^2 = 6y_1$,we get $x_1^2 = 6(\frac{2}{3}x_1) = 4x_1 \implies x_1^2 - 4x_1 = 0 \implies x_1 = 0$ or $x_1 = 4$.
If $x_1 = 4$,$y_1 = \frac{2}{3}(4) = \frac{8}{3}$. Point is $(4, 8/3)$.
Case $2$: $m_n = 1 \implies -\frac{6y_1}{x_1^2} = 1 \implies x_1^2 = -6y_1$.
Substituting into $9y_1^2 = x_1^3$,we get $9y_1^2 = x_1(-6y_1) \implies 9y_1 = -6x_1 \implies y_1 = -\frac{2}{3}x_1$.
Substituting into $x_1^2 = -6y_1$,we get $x_1^2 = -6(-\frac{2}{3}x_1) = 4x_1 \implies x_1 = 4$ or $x_1 = 0$.
If $x_1 = 4$,$y_1 = -\frac{2}{3}(4) = -\frac{8}{3}$. Point is $(4, -8/3)$.
Thus,the points are $(4, 8/3)$ and $(4, -8/3)$.

(B) Given the curve $y = x^2 - 5x + 6$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 5$.
At point $(2, 0)$,the slope of the tangent $m_1 = \left( \frac{dy}{dx} \right)_{(2, 0)} = 2(2) - 5 = -1$.
At point $(3, 0)$,the slope of the tangent $m_2 = \left( \frac{dy}{dx} \right)_{(3, 0)} = 2(3) - 5 = 1$.
Since $m_1 \times m_2 = (-1) \times (1) = -1$,the tangents are perpendicular to each other.
Therefore,the angle between the tangents is $\pi /2$.

(B) The equation of the curve is $y = ax^3$.
The length of the sub-tangent at any point $(x_1, y_1)$ is given by the formula $L = |y_1 \cdot \frac{dx}{dy}|$.
First,find the derivative of the curve with respect to $x$:
$\frac{dy}{dx} = 3ax^2$.
Therefore,$\frac{dx}{dy} = \frac{1}{3ax^2}$.
At the point $(x_1, y_1)$,the length of the sub-tangent is:
$L = |y_1 \cdot \frac{1}{3ax_1^2}|$.
Since the point $(x_1, y_1)$ lies on the curve,we have $y_1 = ax_1^3$.
Substituting $y_1$ into the formula:
$L = |ax_1^3 \cdot \frac{1}{3ax_1^2}| = |\frac{x_1}{3}|$.
Thus,the length of the sub-tangent is $x/3$.

(D) The slope of the tangent to the curve $y = f(x)$ is given by $\frac{dy}{dx}$.
The given line is $2x - 3y = 5$,which can be rewritten as $3y = 2x - 5$ or $y = \frac{2}{3}x - \frac{5}{3}$.
The slope of this line is $m_1 = \frac{2}{3}$.
Since the tangent is perpendicular to this line,the product of their slopes must be $-1$. Let $m_2$ be the slope of the tangent.
$m_1 \times m_2 = -1$
$\frac{2}{3} \times m_2 = -1$
$m_2 = -\frac{3}{2}$.
Therefore,$\frac{dy}{dx} = -\frac{3}{2}$.

(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 $\frac{dy}{dx}$:
$\frac{dy}{dx} (3y^2 - 12) = -6x$
$\frac{dy}{dx} = \frac{-6x}{3y^2 - 12} = \frac{6x}{12 - 3y^2}$.
For the tangent to be parallel to the $y$-axis,the slope $\frac{dy}{dx}$ must be undefined,which means the denominator must be zero:
$12 - 3y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$.
Case $1$: If $y = 2$,then $y^3 + 3x^2 = 12y \implies (2)^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x = \pm \frac{4}{\sqrt{3}}$.
Case $2$: If $y = -2$,then $y^3 + 3x^2 = 12y \implies (-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$.
Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

(B) The slope of the line $4x - 2y + 5 = 0$ is $m = 2$. Since the tangent is parallel to this line,its slope must also be $2$.
Given the curve $y = \sqrt{3x - 2}$,we find the derivative:
$\frac{dy}{dx} = \frac{1}{2\sqrt{3x - 2}} \times 3 = \frac{3}{2\sqrt{3x - 2}}$.
Setting the slope equal to $2$:
$\frac{3}{2\sqrt{3x - 2}} = 2$
$3 = 4\sqrt{3x - 2}$
$9 = 16(3x - 2)$
$9 = 48x - 32$
$48x = 41 \Rightarrow x = \frac{41}{48}$.
Now,find the $y$-coordinate:
$y = \sqrt{3(\frac{41}{48}) - 2} = \sqrt{\frac{41}{16} - \frac{32}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
The point of tangency is $(\frac{41}{48}, \frac{3}{4})$.
The equation of the tangent is $y - y_1 = m(x - x_1)$:
$y - \frac{3}{4} = 2(x - \frac{41}{48})$
$y - \frac{3}{4} = 2x - \frac{41}{24}$
$y = 2x - \frac{41}{24} + \frac{18}{24}$
$y = 2x - \frac{23}{24}$
$24y = 48x - 23$
$48x - 24y - 23 = 0$.

(D) Given the curve equation: $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = a \left( \frac{1}{a} e^{\frac{x}{a}} - \frac{1}{a} e^{-\frac{x}{a}} \right) = e^{\frac{x}{a}} - e^{-\frac{x}{a}}$.
Since the tangent is parallel to the $x$-axis,its slope is $0$:
$\frac{dy}{dx} = 0$.
Therefore,$e^{\frac{x}{a}} - e^{-\frac{x}{a}} = 0$.
$e^{\frac{x}{a}} = e^{-\frac{x}{a}}$.
Taking the natural logarithm on both sides:
$\frac{x}{a} = -\frac{x}{a}$.
$2x = 0$,which implies $x = 0$.

(C) Given the equation of the curve is $\frac{x^2}{4} + \frac{y^2}{25} = 1$.
Differentiating both sides with respect to $x$,we get:
$\frac{2x}{4} + \frac{2y}{25} \frac{dy}{dx} = 0$
Simplifying the expression:
$\frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0$
$\frac{2y}{25} \frac{dy}{dx} = -\frac{x}{2}$
$\frac{dy}{dx} = -\frac{25x}{4y}$
For the tangent to be parallel to the $x$-axis,the slope of the tangent must be $0$,i.e.,$\frac{dy}{dx} = 0$.
Setting the derivative to zero:
$-\frac{25x}{4y} = 0 \implies x = 0$.
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 on the curve where the tangents are parallel to the $x$-axis are $(0, 5)$ and $(0, -5)$.

(B) Given the parametric equations: $x = a(\cos \theta + \theta \sin \theta)$ and $y = a(\sin \theta - \theta \cos \theta)$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(-\sin \theta + \sin \theta + \theta \cos \theta) = a\theta \cos \theta$
$\frac{dy}{d\theta} = a(\cos \theta - \cos \theta + \theta \sin \theta) = a\theta \sin \theta$
Now,find the slope of the tangent:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta$.
The slope of the normal is $m_n = -\frac{1}{dy/dx} = -\frac{1}{\tan \theta} = -\cot \theta$.
Since the slope of the normal is $-\cot \theta = \tan(\theta + \frac{\pi}{2})$,the angle it makes with the $x$-axis is $(\frac{\pi}{2} + \theta)$.
Thus,option $B$ is correct.

(B) Given the curve $y = e^{2x} + x^2$.
First,find the derivative: $\frac{dy}{dx} = 2e^{2x} + 2x$.
At the point $(0, 1)$,the slope of the tangent is $m_t = 2e^0 + 2(0) = 2$.
The slope of the normal $m_n$ is $-\frac{1}{m_t} = -\frac{1}{2}$.
The equation of the normal at $(0, 1)$ is $y - 1 = -\frac{1}{2}(x - 0)$,which simplifies to $x + 2y - 2 = 0$.
To find the intercepts with the axes:
For the $x$-intercept,set $y = 0$: $x + 2(0) - 2 = 0 \implies x = 2$. The point is $(2, 0)$.
For the $y$-intercept,set $x = 0$: $0 + 2y - 2 = 0 \implies y = 1$. The point is $(0, 1)$.
The triangle formed by the normal and the axes has vertices at $(0, 0)$,$(2, 0)$,and $(0, 1)$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$ square unit.

(A) Let $P(h, k)$ be a point on the curve $y = 4x^3 - 2x^5$.
Since $P$ lies on the curve,we have $k = 4h^3 - 2h^5$ ..... $(1)$.
The slope of the tangent at $P(h, k)$ is given by $\frac{dy}{dx} = 12x^2 - 10x^4$.
At $x = h$,the slope $m = 12h^2 - 10h^4$.
The equation of the tangent at $P(h, k)$ is $y - k = (12h^2 - 10h^4)(x - h)$.
Since the tangent passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$:
$0 - k = (12h^2 - 10h^4)(0 - h)$
$-k = -12h^3 + 10h^5$
$k = 12h^3 - 10h^5$ ..... $(2)$.
Equating $(1)$ and $(2)$:
$4h^3 - 2h^5 = 12h^3 - 10h^5$
$8h^5 - 8h^3 = 0$
$8h^3(h^2 - 1) = 0$
This gives $h = 0, h = 1, h = -1$.
For $h = 0$,$k = 4(0)^3 - 2(0)^5 = 0$. Point is $(0, 0)$.
For $h = 1$,$k = 4(1)^3 - 2(1)^5 = 4 - 2 = 2$. Point is $(1, 2)$.
For $h = -1$,$k = 4(-1)^3 - 2(-1)^5 = -4 + 2 = -2$. Point is $(-1, -2)$.
Thus,the points are $(0, 0), (1, 2), (-1, -2)$.

(A) Given the parametric equations: $x = a(\cos t + \log \tan(t/2))$ and $y = a \sin t$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = a(-\sin t + \frac{1}{\tan(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}) = a(-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}) = a(-\sin t + \frac{1}{\sin t}) = a(\frac{1 - \sin^2 t}{\sin t}) = a \frac{\cos^2 t}{\sin t}$.
$\frac{dy}{dt} = a \cos t$.
Now,find the slope of the tangent $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = \frac{\sin t}{\cos t} = \tan t$.
The formula for the length of the tangent is $L = |y \frac{\sqrt{1 + (dy/dx)^2}}{dy/dx}|$.
Substituting the values: $L = |a \sin t \cdot \frac{\sqrt{1 + \tan^2 t}}{\tan t}| = |a \sin t \cdot \frac{\sec t}{\tan t}| = |a \sin t \cdot \frac{1}{\cos t} \cdot \frac{\cos t}{\sin t}| = |a| = a$ (assuming $a > 0$).

(B) Given the curve equation 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,$\frac{dy}{dx} = \frac{a^{n-1}}{n y^{n-1}}$.
The length of the subnormal at any point $(x, y)$ is given by the formula:
$L = \left| y \frac{dy}{dx} \right|$.
Substituting the value of $\frac{dy}{dx}$:
$L = \left| y \cdot \frac{a^{n-1}}{n y^{n-1}} \right| = \left| \frac{a^{n-1}}{n y^{n-2}} \right|$.
For the length of the subnormal to be constant,it must be independent of $y$.
This is possible only if the exponent of $y$ is $0$.
Thus,$n - 2 = 0$,which implies $n = 2$.

(C) Given the curve $y = be^{-x/a}$.
To find the point of contact,we first find the derivative of the curve: $\frac{dy}{dx} = b \cdot e^{-x/a} \cdot (-\frac{1}{a}) = -\frac{b}{a} e^{-x/a}$.
The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $y = -\frac{b}{a}x + b$. The slope of this line is $m = -\frac{b}{a}$.
At the point of contact,the slope of the tangent to the curve must equal the slope of the line:
$-\frac{b}{a} e^{-x/a} = -\frac{b}{a}$.
This implies $e^{-x/a} = 1$,so $-x/a = 0$,which gives $x = 0$.
Substituting $x = 0$ into the curve equation $y = be^{-x/a}$,we get $y = be^0 = b$.
Thus,the point of contact is $(0, b)$.

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

(A) Given the parametric equations of the curve: $x = a \cos^3 t$ and $y = a \sin^3 t$.
First,find the derivatives with respect to $t$:
$\frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$
$\frac{dy}{dt} = 3a \sin^2 t (\cos t) = 3a \sin^2 t \cos t$
Now,find the slope of the tangent $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\sin t}{\cos t} = -\tan t$
The equation of the tangent at point $(a \cos^3 t, a \sin^3 t)$ is:
$y - a \sin^3 t = -\frac{\sin t}{\cos t} (x - a \cos^3 t)$
Multiply both sides by $\cos t$:
$y \cos t - a \sin^3 t \cos t = -x \sin t + a \cos^3 t \sin t$
Rearrange the terms:
$x \sin t + y \cos t = a \sin t \cos^3 t + a \sin^3 t \cos t$
$x \sin t + y \cos t = a \sin t \cos t (\cos^2 t + \sin^2 t)$
Since $\cos^2 t + \sin^2 t = 1$,we get:
$x \sin t + y \cos t = a \sin t \cos t$
Divide both sides by $\sin t \cos t$:
$\frac{x \sin t}{\sin t \cos t} + \frac{y \cos t}{\sin t \cos t} = \frac{a \sin t \cos t}{\sin t \cos t}$
$x \sec t + y \csc t = a$

(C) Given the curve $y = x^3$.
First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 3x^2$.
At the point $P(1, 1)$,the slope of the tangent is $m_t = 3(1)^2 = 3$.
The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = -\frac{1}{3}$.
Using the point-slope form of the equation of a line,$y - y_1 = m_n(x - x_1)$,we get:
$y - 1 = -\frac{1}{3}(x - 1)$.
Multiplying both sides by $3$:
$3(y - 1) = -(x - 1)$.
$3y - 3 = -x + 1$.
Rearranging the terms gives the equation of the normal:
$x + 3y - 4 = 0$.

(D) Given the parametric equations of the curve:
$x = a(1 + \cos \theta)$ and $y = a \sin \theta$.
First,we find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = -a \sin \theta$
$\frac{dy}{d\theta} = a \cos \theta$
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
The slope of the normal at point $\theta$ is the negative reciprocal of the tangent slope:
$m_n = -\frac{1}{-\cot \theta} = \tan \theta$.
The equation of the normal at point $(x_1, y_1) = (a(1 + \cos \theta), a \sin \theta)$ is:
$y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta))$
$y - a \sin \theta = \frac{\sin \theta}{\cos \theta} (x - a - a \cos \theta)$
$y \cos \theta - a \sin \theta \cos \theta = x \sin \theta - a \sin \theta - a \sin \theta \cos \theta$
$y \cos \theta = x \sin \theta - a \sin \theta$
$y \cos \theta = (x - a) \sin \theta$
If we substitute the point $(a, 0)$ into the equation:
$0 \cdot \cos \theta = (a - a) \sin \theta$
$0 = 0$.
Thus,the normal always passes through the point $(a, 0)$.

(D) Given the curve $2y^3 = ax^2 + x^3$. Differentiating with respect to $x$,we get:
$6y^2 \frac{dy}{dx} = 2ax + 3x^2$.
At the point $(a, a)$,the slope of the tangent is:
$\left( \frac{dy}{dx} \right)_{(a, a)} = \frac{2a(a) + 3(a)^2}{6(a)^2} = \frac{5a^2}{6a^2} = \frac{5}{6}$.
The equation of the tangent at $(a, a)$ is:
$y - a = \frac{5}{6}(x - a)$
$6y - 6a = 5x - 5a$
$5x - 6y + a = 0$.
To find the intercepts $p$ and $q$ on the coordinate axes,we write the equation in intercept form:
$5x - 6y = -a$
$\frac{x}{-a/5} + \frac{y}{a/6} = 1$.
Thus,$p = -a/5$ and $q = a/6$.
Given $p^2 + q^2 = 61$,we have:
$\frac{a^2}{25} + \frac{a^2}{36} = 61$
$a^2 \left( \frac{36 + 25}{25 \times 36} \right) = 61$
$a^2 \left( \frac{61}{900} \right) = 61$
$a^2 = 900$
$a = \pm 30$.

(B) Given curves are $y = 1 - ax^2$ and $y = x^2$.
For the first curve,$\frac{dy}{dx} = -2ax$.
For the second curve,$\frac{dy}{dx} = 2x$.
Since the curves intersect orthogonally,the product of their slopes at the point of intersection $(x, y)$ must be $-1$.
So,$(-2ax)(2x) = -1 \implies 4ax^2 = 1 \implies x^2 = \frac{1}{4a}$.
Now,equate the $y$ values of both curves at the intersection point:
$1 - ax^2 = x^2 \implies 1 = (1 + a)x^2$.
Substitute $x^2 = \frac{1}{4a}$ into the equation:
$1 = (1 + a) \left( \frac{1}{4a} \right)$.
$4a = 1 + a \implies 3a = 1 \implies a = 1/3$.

(B) Given the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$.
Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.
At the point $(x_1, y_1)$,the slope of the tangent is $m = -\sqrt{\frac{y_1}{x_1}}$.
The equation of the tangent is $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1)$.
Multiplying by $\sqrt{x_1}$,we get $y\sqrt{x_1} - y_1\sqrt{x_1} = -x\sqrt{y_1} + x_1\sqrt{y_1}$.
Rearranging terms,$x\sqrt{y_1} + y\sqrt{x_1} = x_1\sqrt{y_1} + y_1\sqrt{x_1}$.
Since $(x_1, y_1)$ lies on the curve,$\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$.
Dividing the equation $x\sqrt{y_1} + y\sqrt{x_1} = \sqrt{x_1}\sqrt{y_1}(\sqrt{x_1} + \sqrt{y_1})$ by $\sqrt{x_1}\sqrt{y_1}$,we get $\frac{x}{\sqrt{x_1}} + \frac{y}{\sqrt{y_1}} = \sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$.

(A) Given the curve $y = x^2 - 5x + 6$.
First,find the derivative: $\frac{dy}{dx} = 2x - 5$.
At point $(2, 0)$,the slope of the tangent $m_1 = \left. \frac{dy}{dx} \right|_{(2,0)} = 2(2) - 5 = -1$.
At point $(3, 0)$,the slope of the tangent $m_2 = \left. \frac{dy}{dx} \right|_{(3,0)} = 2(3) - 5 = 1$.
Let $\theta$ be the angle between the tangents. The formula is $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{0} \right| = \infty$.
Since $\tan \theta = \infty$,the angle $\theta = \frac{\pi}{2}$.

(D) For the curve $y = x^2$,the slope $m_1$ is given by $\frac{dy}{dx} = 2x$.
At the point $(1, 1)$,$m_1 = 2(1) = 2$.
For the curve $x = y^2$,differentiating with respect to $x$ gives $1 = 2y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2y}$.
At the point $(1, 1)$,$m_2 = \frac{1}{2(1)} = \frac{1}{2}$.
The angle $\theta$ between the two 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,$\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

(C) The slope of the normal to the curve at point $(3, 4)$ is given by $m_n = \tan(3\pi / 4) = -1$.
The slope of the tangent to the curve at point $(3, 4)$ is $m_t = -1 / m_n$.
Substituting the value of $m_n$,we get $m_t = -1 / (-1) = 1$.
Since the slope of the tangent at $x = 3$ is $f'(3)$,we have $f'(3) = 1$.

(C) The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.
Given $y = \sqrt{4x - 3} - 1$,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} \times 4 = \frac{2}{\sqrt{4x - 3}}$.
We are given that the slope of the tangent is $\frac{2}{3}$.
Equating the derivative to the given slope:
$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}$
$\sqrt{4x - 3} = 3$
Squaring both sides:
$4x - 3 = 9$
$4x = 12$
$x = 3$.
Now,substitute $x = 3$ into the original equation of the curve to find $y$:
$y = \sqrt{4(3) - 3} - 1$
$y = \sqrt{12 - 3} - 1$
$y = \sqrt{9} - 1$
$y = 3 - 1 = 2$.
Thus,the required point is $(3, 2)$.

(A) Given the curve $y = e^x$,the slope of the tangent at $(c, e^c)$ is given by $\frac{dy}{dx} = e^x$.
At $x = c$,the slope $m_1 = e^c$.
The equation of the tangent at $(c, e^c)$ is $y - e^c = e^c(x - c)$,which simplifies to $y = e^c(x - c + 1)$.
Now,the slope of the line joining $(c - 1, e^{c-1})$ and $(c + 1, e^{c+1})$ is $m_2 = \frac{e^{c+1} - e^{c-1}}{(c+1) - (c-1)} = \frac{e^{c-1}(e^2 - 1)}{2}$.
The equation of this secant line is $y - e^{c-1} = m_2(x - (c - 1))$.
To find the intersection,we set the $y$-values equal: $e^c(x - c + 1) = e^{c-1}(x - c + 1) + e^{c-1}$.
Solving for $x$,we find that the $x$-coordinate of the intersection is $c - 1 + \frac{2}{e^2 - 1} \cdot e$.
Since $e \approx 2.718$,$e^2 - 1 \approx 6.389$.
Thus,$x = c - 1 + \frac{2e}{e^2 - 1} < c - 1 + 1 = c$.
Therefore,the intersection point has an $x$-coordinate less than $c$.

(B) Let the point of intersection of the two curves be $(x_1, y_1)$.
The equations of the curves are $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$.
Differentiating both equations with respect to $x$:
For $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$,we get $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{a^2y}$.
For $y^3 = 16x$,we get $3y^2 \frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{16}{3y^2}$.
Let $m_1$ and $m_2$ be the slopes of the tangents at $(x_1, y_1)$.
$m_1 = -\frac{4x_1}{a^2y_1}$ and $m_2 = \frac{16}{3y_1^2}$.
Since the curves intersect at right angles,$m_1 m_2 = -1$.
$(-\frac{4x_1}{a^2y_1}) \times (\frac{16}{3y_1^2}) = -1$.
$\frac{64x_1}{3a^2y_1^3} = 1$.
Since $(x_1, y_1)$ lies on $y^3 = 16x$,we have $y_1^3 = 16x_1$.
Substituting this into the equation: $\frac{64x_1}{3a^2(16x_1)} = 1$.
$\frac{64x_1}{48a^2x_1} = 1 \Rightarrow \frac{4}{3a^2} = 1$.
Therefore,$a^2 = 4/3$.

(C) Given the curve $y = 2 \cos x$.
At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2}$.
So,the point of tangency is $\left( \frac{\pi}{4}, \sqrt{2} \right)$.
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 passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
Substituting the values,we get $y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$.

(B) Let the curve be $y = f(x)$. The length of the subtangent is given by $|y_1 / (dy/dx)|$ and the length of the subnormal is given by $|y_1 (dy/dx)|$.
Given that the length of the subtangent equals the length of the subnormal:
$|y_1 / (dy/dx)| = |y_1 (dy/dx)|$
Assuming $y_1 \neq 0$,we have $1 / |dy/dx| = |dy/dx|$,which implies $(dy/dx)^2 = 1$,so $dy/dx = \pm 1$.
The length of the tangent is given by the formula $L_t = |y_1| \sqrt{1 + (dx/dy)^2}$.
Since $dy/dx = \pm 1$,then $dx/dy = \pm 1$.
Substituting this into the formula:
$L_t = |y_1| \sqrt{1 + (\pm 1)^2} = |y_1| \sqrt{1 + 1} = \sqrt{2} |y_1|$.
Thus,the length of the tangent is $\sqrt{2} y_1$ (assuming $y_1 > 0$).

(C) Given the curve equation: $y = (x + 1)(x - 3)$.
To find the points where the curve meets the $x$-axis,set $y = 0$:
$0 = (x + 1)(x - 3)$
This gives $x = -1$ and $x = 3$.
So,the points of intersection are $(-1, 0)$ and $(3, 0)$.
Now,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$y = x^2 - 2x - 3$
$\frac{dy}{dx} = 2x - 2$
At point $(-1, 0)$:
Slope $m_1 = 2(-1) - 2 = -2 - 2 = -4$.
At point $(3, 0)$:
Slope $m_2 = 2(3) - 2 = 6 - 2 = 4$.
Thus,the slopes of the tangents at these points are $\pm 4$.

(A) Given the equation of the hyperbola: $2x^2 - 3y^2 = 6$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(2x^2) - \frac{d}{dx}(3y^2) = \frac{d}{dx}(6)$
$4x - 6y \cdot \frac{dy}{dx} = 0$
Rearranging to solve for $\frac{dy}{dx}$:
$6y \cdot \frac{dy}{dx} = 4x$
$\frac{dy}{dx} = \frac{4x}{6y} = \frac{2x}{3y}$
Now,substitute the point $(3, 2)$ into the derivative to find the slope $m$:
$m = \left( \frac{dy}{dx} \right)_{(3, 2)} = \frac{2(3)}{3(2)} = \frac{6}{6} = 1$
Thus,the slope of the tangent at $(3, 2)$ is $1$.

(C) To find the angle between the curves $y = \sin x$ and $y = \cos x$,we first find their point of intersection by setting $\sin x = \cos x$.
This gives $\tan x = 1$,so $x = \pi /4$.
At $x = \pi /4$,$y = \sin(\pi /4) = 1/\sqrt{2}$. The point of intersection is $(\pi /4, 1/\sqrt{2})$.
Now,find the slopes of the tangents at this point:
For $y = \sin x$,$dy/dx = \cos x$. At $x = \pi /4$,$m_1 = \cos(\pi /4) = 1/\sqrt{2}$.
For $y = \cos x$,$dy/dx = -\sin x$. At $x = \pi /4$,$m_2 = -\sin(\pi /4) = -1/\sqrt{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = |(m_1 - m_2) / (1 + m_1 m_2)|$.
$\tan \theta = |(1/\sqrt{2} - (-1/\sqrt{2})) / (1 + (1/\sqrt{2})(-1/\sqrt{2}))|$
$\tan \theta = |(2/\sqrt{2}) / (1 - 1/2)| = |\sqrt{2} / (1/2)| = 2\sqrt{2}$.
Therefore,$\theta = \tan^{-1}(2\sqrt{2})$.

(C) Given the curve $y = 2x^2 + 3x - 2$.
First,find the derivative: $\frac{dy}{dx} = 4x + 3$.
At the point $(1, 3)$,the slope of the tangent is $m = 4(1) + 3 = 7$.
The equation of the tangent line at $(1, 3)$ is given by $y - 3 = 7(x - 1)$,which simplifies to $y - 3 = 7x - 7$,or $7x - y = 4$.
To find the $x$-intercept,set $y = 0$: $7x = 4 \implies x = 4/7$.
To find the $y$-intercept,set $x = 0$: $-y = 4 \implies y = -4$.
Thus,the intercepts on the coordinate axes are $4/7$ and $-4$.

(A) Given curves are $y = 4x^2$ and $y = x^2$.
To find the point of intersection,set $4x^2 = x^2$,which implies $3x^2 = 0$,so $x = 0$. Thus,the point of intersection is $(0, 0)$.
Now,find the slopes of the tangents at $(0, 0)$:
For $y = 4x^2$,$\frac{dy}{dx} = 8x$. At $x = 0$,$m_1 = 8(0) = 0$.
For $y = x^2$,$\frac{dy}{dx} = 2x$. At $x = 0$,$m_2 = 2(0) = 0$.
Since both slopes are $0$,the tangents are horizontal at the origin.
The angle of intersection $\theta$ is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{0 - 0}{1 + 0}| = 0$.
Therefore,$\theta = 0^\circ$.

(D) Given the curve $y = \frac{1}{x^2 - 2x + 3}$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{-1}{(x^2 - 2x + 3)^2} \cdot (2x - 2) = \frac{-(2x - 2)}{(x^2 - 2x + 3)^2}$.
Since the slope of the tangent is $0$,we set $\frac{dy}{dx} = 0$:
$\frac{-(2x - 2)}{(x^2 - 2x + 3)^2} = 0 \implies 2x - 2 = 0 \implies x = 1$.
Now,substitute $x = 1$ into the original equation to find the $y$-coordinate:
$y = \frac{1}{(1)^2 - 2(1) + 3} = \frac{1}{1 - 2 + 3} = \frac{1}{2}$.
The point of tangency is $(1, 1/2)$.
The equation of the tangent line with slope $m = 0$ passing through $(1, 1/2)$ is:
$y - \frac{1}{2} = 0(x - 1) \implies y = \frac{1}{2} \implies 2y - 1 = 0$.

(C) Let the slope of the tangent at a point $(x, y)$ on the curve be $m = \frac{dy}{dx}$.
The length of the subtangent is given by $|\frac{y}{m}| = |y \frac{dx}{dy}|$.
The length of the subnormal is given by $|my| = |y \frac{dy}{dx}|$.
Now,consider the expression $\sqrt{\frac{\text{subnormal}}{\text{subtangent}}}$.
Substituting the formulas,we get $\sqrt{\frac{|y \frac{dy}{dx}|}{|y \frac{dx}{dy}|}} = \sqrt{|\frac{dy}{dx} \cdot \frac{dy}{dx}|} = \sqrt{(\frac{dy}{dx})^2} = |\frac{dy}{dx}|$.
Thus,$\sqrt{\frac{\text{subnormal}}{\text{subtangent}}} = |\frac{dy}{dx}|$,which is the magnitude of the slope of the tangent at that point.

(B) Given the curve equation: $y = \frac{a}{2}(e^{x/a} + e^{-x/a}) = a \cosh(\frac{x}{a})$.
Find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = a \cdot \frac{1}{a} \sinh(\frac{x}{a}) = \sinh(\frac{x}{a})$.
The formula for the length of the normal is $L_n = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.
Substitute the values:
$L_n = y \sqrt{1 + \sinh^2(\frac{x}{a})}$.
Using the identity $1 + \sinh^2(\theta) = \cosh^2(\theta)$:
$L_n = y \sqrt{\cosh^2(\frac{x}{a})} = y \cosh(\frac{x}{a})$.
Since $y = a \cosh(\frac{x}{a})$,we have $\cosh(\frac{x}{a}) = \frac{y}{a}$.
Therefore,$L_n = y \cdot (\frac{y}{a}) = \frac{y^2}{a}$.

(C) To find the intersection points,set $4 - x^2 = x^2$,which gives $2x^2 = 4$,so $x^2 = 2$,$x = \pm \sqrt{2}$.
For $x = \pm \sqrt{2}$,$y = 2$. Thus,the points of intersection are $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.
For the curve $y = 4 - x^2$,the slope $m_1 = \frac{dy}{dx} = -2x$.
For the curve $y = x^2$,the slope $m_2 = \frac{dy}{dx} = 2x$.
At the point $(\sqrt{2}, 2)$,$m_1 = -2\sqrt{2}$ and $m_2 = 2\sqrt{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{-2\sqrt{2} - 2\sqrt{2}}{1 + (-2\sqrt{2})(2\sqrt{2})} \right| = \left| \frac{-4\sqrt{2}}{1 - 8} \right| = \left| \frac{-4\sqrt{2}}{-7} \right| = \frac{4\sqrt{2}}{7}$.
Therefore,$\theta = \tan^{-1}\left( \frac{4\sqrt{2}}{7} \right)$.

(A) Given the curve equation: $\sqrt{x} + \sqrt{y} = \sqrt{a}$.
Differentiating with respect to $x$: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
$\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$.
The equation of the tangent at point $(x_1, y_1)$ is $Y - y_1 = -\frac{\sqrt{y_1}}{\sqrt{x_1}}(X - x_1)$.
$Y\sqrt{x_1} - y_1\sqrt{x_1} = -X\sqrt{y_1} + x_1\sqrt{y_1}$.
$X\sqrt{y_1} + Y\sqrt{x_1} = x_1\sqrt{y_1} + y_1\sqrt{x_1} = \sqrt{x_1 y_1}(\sqrt{x_1} + \sqrt{y_1})$.
Since $\sqrt{x_1} + \sqrt{y_1} = \sqrt{a}$,we have $X\sqrt{y_1} + Y\sqrt{x_1} = \sqrt{a}\sqrt{x_1 y_1}$.
Dividing by $\sqrt{a}\sqrt{x_1 y_1}$: $\frac{X}{\sqrt{a}\sqrt{x_1}} + \frac{Y}{\sqrt{a}\sqrt{y_1}} = 1$.
The intercepts on the axes are $X_{int} = \sqrt{a}\sqrt{x_1}$ and $Y_{int} = \sqrt{a}\sqrt{y_1}$.
The sum of the intercepts is $\sqrt{a}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{a}(\sqrt{a}) = a$.

(A) The given line is $2x - 2y + 3 = 0$,which can be written as $y = x + \frac{3}{2}$. The slope of this line is $m = 1$.
Since the normal is parallel to this line,the slope of the normal is $m_n = 1$.
The slope of the tangent to the curve $y = x \log x$ is $\frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$.
The slope of the normal is $m_n = -\frac{1}{dy/dx} = -\frac{1}{\log x + 1}$.
Equating the slopes: $-\frac{1}{\log x + 1} = 1 \implies \log x + 1 = -1 \implies \log x = -2 \implies x = e^{-2}$.
For $x = e^{-2}$,the value of $y$ is $y = e^{-2} \log(e^{-2}) = e^{-2} (-2) = -2e^{-2}$.
The point of contact is $(e^{-2}, -2e^{-2})$.
The equation of the normal is $y - y_1 = m_n(x - x_1)$.
$y - (-2e^{-2}) = 1(x - e^{-2})$.
$y + 2e^{-2} = x - e^{-2}$.
$x - y = 3e^{-2}$.

(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 equation $y = x^3 - x$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(x^3 - x) = 3x^2 - 1$.
To find the slope of the tangent at $x = 2$,we evaluate the derivative at $x = 2$:
$\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^2 - 1$.
$= 3(4) - 1$.
$= 12 - 1 = 11$.
Therefore,the slope of the tangent at $x = 2$ is $11$.

(C) Given the equation of the curve is $y = (x - 1)(x - 2) = x^2 - 3x + 2$.
To find the slope of the tangent,we differentiate with respect to $x$:
$\frac{dy}{dx} = 2x - 3$.
Now,evaluate the slope at the point $(1, 0)$:
$m = \left( \frac{dy}{dx} \right)_{(1, 0)} = 2(1) - 3 = -1$.
The slope of the tangent is equal to $\tan(\psi)$,where $\psi$ is the angle of inclination with the $x$-axis.
$\tan(\psi) = -1$.
Since $\tan(\psi) = -1$,the angle $\psi = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(NONE) Given the curve $y = x^2 - x + 4$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 1$.
At point $P(1, 4)$,the slope of the tangent is $m = \left(\frac{dy}{dx}\right)_{(1, 4)} = 2(1) - 1 = 1$.
The equation of the tangent at $(1, 4)$ is $y - 4 = 1(x - 1)$,which simplifies to $y = x + 3$.
For point $A$ on the $X$-axis,set $y = 0$: $0 = x + 3 \implies x = -3$. So,$A = (-3, 0)$.
The slope of the normal at $(1, 4)$ is $m' = -\frac{1}{m} = -1$.
The equation of the normal at $(1, 4)$ is $y - 4 = -1(x - 1)$,which simplifies to $y = -x + 5$.
For point $B$ on the $X$-axis,set $y = 0$: $0 = -x + 5 \implies x = 5$. So,$B = (5, 0)$.
The coordinates are $P(1, 4)$,$A(-3, 0)$,and $B(5, 0)$.
The base $AB$ of $\Delta PAB$ lies on the $X$-axis. Length $AB = |5 - (-3)| = 8$.
The height of $\Delta PAB$ is the $y$-coordinate of $P$,which is $4$.
Area of $\Delta PAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 4 = 16$ square units.
Note: The provided solution in the input was incorrect; the correct calculation yields $16$.

(A) Let the curve be $y = f(x)$.
At any point $(x, y)$ on the curve,let $m = \frac{dy}{dx}$ be the slope of the tangent.
The length of the tangent is given by $L_T = |y| \sqrt{1 + \frac{1}{m^2}} = |y| \frac{\sqrt{1 + m^2}}{|m|}$.
The length of the normal is given by $L_N = |y| \sqrt{1 + m^2}$.
The length of the subtangent is $S_T = |\frac{y}{m}|$.
The length of the subnormal is $S_N = |ym|$.
Now,consider the ratio $\frac{(L_N)^2}{(L_T)^2} = \frac{y^2(1 + m^2)}{y^2 \frac{1 + m^2}{m^2}} = m^2$.
Also,the ratio of subnormal to subtangent is $\frac{S_N}{S_T} = \frac{|ym|}{|y/m|} = |m^2| = m^2$.
Therefore,$\frac{(L_N)^2}{(L_T)^2} = \frac{S_N}{S_T}$.

(D) The slope of the tangent to the curve $y = f(x)$ at $x = 3$ is given by $f'(3)$.
The slope of the normal to the curve at the point $(3, 4)$ is given by $m_n = -\frac{1}{f'(3)}$.
Given that the normal makes an angle of $\theta = 3\pi /4$ with the $x$-axis,the slope of the normal is $m_n = \tan(3\pi /4)$.
We know that $\tan(3\pi /4) = -1$.
Therefore,$-\frac{1}{f'(3)} = -1$.
Multiplying both sides by $-1$,we get $\frac{1}{f'(3)} = 1$.
Thus,$f'(3) = 1$.

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

(D) Given the curve $x^3 - y^2 = 0$.
By differentiating with respect to $x$,we get $3x^2 - 2y \frac{dy}{dx} = 0$.
So,$\frac{dy}{dx} = \frac{3x^2}{2y}$.
At the point $(m^2, -m^3)$,the slope of the tangent is $\left( \frac{dy}{dx} \right)_{(m^2, -m^3)} = \frac{3(m^2)^2}{2(-m^3)} = \frac{3m^4}{-2m^3} = -\frac{3}{2}m$.
The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{-3m/2} = \frac{2}{3m}$.
The equation of the normal at $(m^2, -m^3)$ is $y - (-m^3) = \frac{2}{3m}(x - m^2)$.
$y + m^3 = \frac{2}{3m}x - \frac{2m^2}{3m} \Rightarrow y = \frac{2}{3m}x - \frac{2m}{3} - m^3$.
Comparing this with the given equation $y = 3mx - 4m^3$,we equate the coefficients of $x$:
$3m = \frac{2}{3m} \Rightarrow 9m^2 = 2 \Rightarrow m^2 = \frac{2}{9}$.
Also,comparing the constant terms: $-4m^3 = -m^3 - \frac{2m}{3} \Rightarrow 3m^3 = \frac{2m}{3} \Rightarrow 9m^2 = 2 \Rightarrow m^2 = \frac{2}{9}$ (for $m \neq 0$).

(D) Given the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$.
Differentiating with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$.
Thus,$\left( \frac{dy}{dx} \right)_{(x_1, y_1)} = -\sqrt{\frac{y_1}{x_1}}$.
The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1)$.
For the $x$-intercept,set $y = 0$: $-y_1 = -\sqrt{\frac{y_1}{x_1}}(x - x_1) \implies x - x_1 = y_1 \sqrt{\frac{x_1}{y_1}} = \sqrt{x_1 y_1}$.
So,$x = x_1 + \sqrt{x_1 y_1} = \sqrt{x_1}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{x_1} \sqrt{a} = \sqrt{ax_1}$.
For the $y$-intercept,set $x = 0$: $y - y_1 = -\sqrt{\frac{y_1}{x_1}}(-x_1) = \sqrt{x_1 y_1}$.
So,$y = y_1 + \sqrt{x_1 y_1} = \sqrt{y_1}(\sqrt{y_1} + \sqrt{x_1}) = \sqrt{y_1} \sqrt{a} = \sqrt{ay_1}$.
The sum of the intercepts is $\sqrt{ax_1} + \sqrt{ay_1} = \sqrt{a}(\sqrt{x_1} + \sqrt{y_1}) = \sqrt{a} \cdot \sqrt{a} = a$.

(A) Let the required point be $(x_1, y_1)$.
The equation of the curve is $9y^2 = x^3$.
Since $(x_1, y_1)$ lies on the curve,we have $9y_1^2 = x_1^3$ ... $(i)$.
Differentiating $9y^2 = x^3$ with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,which implies $\frac{dy}{dx} = \frac{x^2}{6y}$.
The slope of the tangent at $(x_1, y_1)$ is $m_t = \frac{x_1^2}{6y_1}$.
The slope of the normal at $(x_1, y_1)$ 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 $\pm 1$.
Therefore,$-\frac{6y_1}{x_1^2} = \pm 1$,which implies $x_1^2 = \mp 6y_1$.
Squaring both sides,$x_1^4 = 36y_1^2$. Substituting $y_1^2 = \frac{x_1^3}{9}$ from $(i)$,we get $x_1^4 = 36 \left( \frac{x_1^3}{9} \right) = 4x_1^3$.
This gives $x_1^3(x_1 - 4) = 0$,so $x_1 = 0$ or $x_1 = 4$.
If $x_1 = 0$,then $y_1 = 0$. However,the normal at $(0,0)$ is not well-defined in a way that creates equal intercepts (it would pass through the origin).
If $x_1 = 4$,then $9y_1^2 = 4^3 = 64$,so $y_1^2 = \frac{64}{9}$,which gives $y_1 = \pm \frac{8}{3}$.
Thus,the required points are $(4, 8/3)$ and $(4, -8/3)$.