Tangent and Normal Questions in English

(D) The given curve is $y^2(x-a)=x^2(x+a)$.
For a tangent to be parallel to the $X$-axis,the slope $\frac{dy}{dx}$ must be $0$.
Differentiating $y^2(x-a)=x^2(x+a)$ with respect to $x$ using implicit differentiation:
$2y \frac{dy}{dx}(x-a) + y^2(1) = 2x(x+a) + x^2(1)$
Setting $\frac{dy}{dx} = 0$,we get:
$y^2 = 2x^2 + 2ax + x^2 = 3x^2 + 2ax$
Substitute $y^2 = \frac{x^2(x+a)}{x-a}$ into the equation:
$\frac{x^2(x+a)}{x-a} = 3x^2 + 2ax$
Since $x=0$ is a solution (giving $y=0$),we check for other solutions by dividing by $x$ $(x \neq 0)$:
$\frac{x(x+a)}{x-a} = 3x + 2a$
$x^2 + ax = (3x + 2a)(x - a)$
$x^2 + ax = 3x^2 - 3ax + 2ax - 2a^2$
$x^2 + ax = 3x^2 - ax - 2a^2$
$2x^2 - 2ax - 2a^2 = 0$
$x^2 - ax - a^2 = 0$
The roots are $x = \frac{a \pm \sqrt{a^2 - 4(1)(-a^2)}}{2} = \frac{a \pm a\sqrt{5}}{2}$.
For these values of $x$,$y^2 = 3x^2 + 2ax$. Since $x^2 - ax - a^2 = 0$,$x^2 = ax + a^2$.
$y^2 = 3(ax + a^2) + 2ax = 5ax + 3a^2$.
For $x = \frac{a(1+\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1+\sqrt{5})}{2}) + 3a^2 > 0$,giving two values of $y$.
For $x = \frac{a(1-\sqrt{5})}{2}$,$y^2 = 5a(\frac{a(1-\sqrt{5})}{2}) + 3a^2 = \frac{5a^2 - 5a^2\sqrt{5} + 6a^2}{2} = \frac{11-5\sqrt{5}}{2}a^2 < 0$,which is impossible.
Thus,there are $2$ tangents.

(A) The equation of the tangent is $2y = 3x - 1$,which can be written as $y = \frac{3}{2}x - \frac{1}{2}$.
Comparing this with $y = mx + c$,the slope of the tangent is $m = \frac{3}{2}$.
Given the curve $y^2 = ax^3 + b$,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = 3ax^2 \implies \frac{dy}{dx} = \frac{3ax^2}{2y}$.
At the point $(1, 1)$,the slope is $\frac{dy}{dx} = \frac{3a(1)^2}{2(1)} = \frac{3a}{2}$.
Equating the slopes: $\frac{3a}{2} = \frac{3}{2} \implies a = 1$.
Since the point $(1, 1)$ lies on the curve $y^2 = ax^3 + b$,we substitute $x=1, y=1, a=1$:
$1^2 = 1(1)^3 + b \implies 1 = 1 + b \implies b = 0$.
Thus,$(a, b) = (1, 0)$.

(B) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$.
The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = \cos x$. At $(h, k)$,the slope is $\cos h$.
The equation of the tangent at $(h, k)$ is $y - k = \cos h(x - h)$.
Since the tangent passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$:
$0 - k = \cos h(0 - h) \Rightarrow -k = -h \cos h \Rightarrow k = h \cos h$.
From $k = \sin h$,we have $\cos h = \frac{k}{h}$.
Using the identity $\sin^2 h + \cos^2 h = 1$,we substitute $\sin h = k$ and $\cos h = \frac{k}{h}$:
$k^2 + \left(\frac{k}{h}\right)^2 = 1
k^2 + \frac{k^2}{h^2} = 1
h^2 k^2 + k^2 = h^2
h^2 - k^2 = h^2 k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 = x^2 y^2$.

(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$
Let $P(h, k)$ be a point on the curve.
Differentiating equation $(i)$ with respect to $x$,we get $6 y \frac{dy}{dx} = 12 x^2$,which implies $\frac{dy}{dx} = \frac{2 x^2}{y}$.
At point $(h, k)$,the slope $m = \frac{2 h^2}{k}$.
The length of the subtangent $T = \left| \frac{k}{m} \right| = \left| \frac{k}{2 h^2 / k} \right| = \frac{k^2}{2 h^2}$.
Since $3 k^2 = 4 h^3$,we have $k^2 = \frac{4}{3} h^3$.
Substituting this into $T$,we get $T = \frac{4 h^3}{3(2 h^2)} = \frac{2}{3} h$.
The length of the subnormal $N = |k m| = \left| k \cdot \frac{2 h^2}{k} \right| = 2 h^2$.
We need to find the relation between $T$ and $N$.
From $N = 2 h^2$,we have $h^2 = \frac{N}{2}$,so $h = \sqrt{\frac{N}{2}}$.
Then $T = \frac{2}{3} \sqrt{\frac{N}{2}}$.
Squaring both sides,$T^2 = \frac{4}{9} \cdot \frac{N}{2} = \frac{2 N}{9}$.
Thus,$9 T^2 = 2 N$,which can be written as $(3 T)^2 = 2 N$.
Comparing this with $(\beta T)^2 = 2 N$,we find $\beta = 3$.
Therefore,the correct option is $C$.

(C) Let $f(x) = x^3$. The slope of the chord joining the points $(-1, -1)$ and $(2, 8)$ is given by $m = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - (-1)}{3} = \frac{9}{3} = 3$.
Since the tangent line is parallel to this chord,the derivative $f'(x)$ at the point must be equal to the slope of the chord.
$f'(x) = \frac{d}{dx}(x^3) = 3x^2$.
Setting $f'(x) = 3$,we get $3x^2 = 3$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.
For $x = 1$,$y = (1)^3 = 1$,giving the point $(1, 1)$.
For $x = -1$,$y = (-1)^3 = -1$,giving the point $(-1, -1)$.
Since $(-1, -1)$ is one of the endpoints of the chord,the point on the curve between the given points is $(1, 1)$.

(D) To find the point of intersection of the curves $x^2+4y=0$ and $xy=2$:
Substitute $y = \frac{2}{x}$ into $x^2+4y=0$:
$x^2 + 4(\frac{2}{x}) = 0$
$x^2 + \frac{8}{x} = 0$
$x^3 + 8 = 0 \Rightarrow x^3 = -8 \Rightarrow x = -2$.
For $x = -2$,$y = \frac{2}{-2} = -1$. So,the point of intersection is $(-2, -1)$.
Slope of the first curve $x^2+4y=0$:
Differentiating with respect to $x$: $2x + 4\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{2}$.
At $x = -2$,$m_1 = -\frac{-2}{2} = 1$.
Slope of the second curve $xy=2$:
Differentiating with respect to $x$: $y + x\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
At $(-2, -1)$,$m_2 = -\frac{-1}{-2} = -\frac{1}{2}$.
The angle $\theta$ between the curves is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} \right| = \left| \frac{3/2}{1/2} \right| = 3$.

(C) Given the curve equation is $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$.
To find the slope of the tangent at $(a, b)$,we differentiate both sides with respect to $x$:
$n\left(\frac{x}{a}\right)^{n-1} \cdot \frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$.
At the point $(a, b)$,we substitute $x = a$ and $y = b$:
$n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$.
$\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0$.
$\frac{n}{b} \cdot \frac{dy}{dx} = -\frac{n}{a} \implies \frac{dy}{dx} = -\frac{b}{a}$.
The equation of the tangent line at $(a, b)$ with slope $m = -\frac{b}{a}$ is given by $y - b = m(x - a)$.
$y - b = -\frac{b}{a}(x - a)$.
$a(y - b) = -b(x - a)$.
$ay - ab = -bx + ab$.
$bx + ay = 2ab$.
Dividing both sides by $ab$,we get $\frac{bx}{ab} + \frac{ay}{ab} = \frac{2ab}{ab}$.
$\frac{x}{a} + \frac{y}{b} = 2$.
Thus,the correct option is $C$.

(B) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$.
Parametric equations are $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
The derivative $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta$.
The slope of the tangent is $m_T = -\tan \theta$.
The slope of the normal is $m_N = \frac{-1}{m_T} = \cot \theta$.
Given that the normal makes an angle $\phi$ with the $X$-axis,its slope is $\tan \phi$.
Thus,$\tan \phi = \cot \theta = \tan(\frac{\pi}{2} - \theta)$,which implies $\phi = \frac{\pi}{2} - \theta$,or $\theta = \frac{\pi}{2} - \phi$.
Substituting $\theta$ into the parametric coordinates: $x = a \cos^3(\frac{\pi}{2} - \phi) = a \sin^3 \phi$ and $y = a \sin^3(\frac{\pi}{2} - \phi) = a \cos^3 \phi$.
The equation of the normal at $(a \sin^3 \phi, a \cos^3 \phi)$ with slope $\tan \phi$ is:
$y - a \cos^3 \phi = \tan \phi (x - a \sin^3 \phi)$
$y \cos \phi - a \cos^4 \phi = x \sin \phi - a \sin^4 \phi$
$y \cos \phi - x \sin \phi = a (\cos^4 \phi - \sin^4 \phi)$
$y \cos \phi - x \sin \phi = a (\cos^2 \phi - \sin^2 \phi)(\cos^2 \phi + \sin^2 \phi)$
$y \cos \phi - x \sin \phi = a \cos 2 \phi$.

(A) Given curves are:
$x^2 = 3y \quad ...(i)$
$x^2 + y^2 = 4 \quad ...(ii)$
Substituting $x^2 = 3y$ into equation $(ii)$,we get:
$3y + y^2 = 4$
$y^2 + 3y - 4 = 0$
$(y + 4)(y - 1) = 0$
Since $x^2 = 3y$,$y$ must be non-negative,so $y = 1$.
Substituting $y = 1$ into $x^2 = 3y$,we get $x^2 = 3$,so $x = \pm \sqrt{3}$.
Thus,the points of intersection are $(\sqrt{3}, 1)$ and $(-\sqrt{3}, 1)$.
Now,differentiating the curves with respect to $x$:
For $(i)$,$2x = 3 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{3}$.
For $(ii)$,$2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$.
At the point $(\sqrt{3}, 1)$:
$m_1 = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$
$m_2 = -\frac{\sqrt{3}}{1} = -\sqrt{3}$
The angle $\theta$ between the curves is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{2}{\sqrt{3}} - (-\sqrt{3})}{1 + (\frac{2}{\sqrt{3}})(-\sqrt{3})} \right|$
$\tan \theta = \left| \frac{\frac{2 + 3}{\sqrt{3}}}{1 - 2} \right| = \left| \frac{5/\sqrt{3}}{-1} \right| = \frac{5}{\sqrt{3}}$
Therefore,$\theta = \tan^{-1} \left( \frac{5}{\sqrt{3}} \right)$.

(D) Given curves are $C_1: x^2 y = 1$ and $C_2: y(x^2 + 1) = 2$.
First,find the intersection points by substituting $y = 1/x^2$ into the second equation: $(1/x^2)(x^2 + 1) = 2 \implies 1 + 1/x^2 = 2 \implies 1/x^2 = 1 \implies x^2 = 1 \implies x = \pm 1$.
For $x = 1$,$y = 1$. For $x = -1$,$y = 1$. Intersection points are $(1, 1)$ and $(-1, 1)$.
For $C_1$,$y = x^{-2} \implies dy/dx = -2x^{-3} = -2/x^3$. At $(1, 1)$,$m_1 = -2$.
For $C_2$,$y = 2/(x^2 + 1) \implies dy/dx = -2(2x)/(x^2 + 1)^2 = -4x/(x^2 + 1)^2$. At $(1, 1)$,$m_2 = -4(1)/(1+1)^2 = -4/4 = -1$.
The angle $\theta$ between the curves is given by $\tan \theta = |(m_1 - m_2) / (1 + m_1 m_2)|$.
$\tan \theta = |(-2 - (-1)) / (1 + (-2)(-1))| = |-1 / (1 + 2)| = |-1/3| = 1/3$.
Thus,$\theta = \operatorname{Tan}^{-1}(1/3)$.

(D) Given curves are:
$x^2+p y^2=1$ $(i)$
$q x^2+y^2=1$ (ii)
Differentiating $(i)$ with respect to $x$:
$2x + 2py \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_1 = -\frac{x}{py}$
Differentiating (ii) with respect to $x$:
$2qx + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = -\frac{qx}{y}$
Since the curves are orthogonal,$m_1 \cdot m_2 = -1$:
$(-\frac{x}{py}) \cdot (-\frac{qx}{y}) = -1$
$\frac{qx^2}{py^2} = -1 \implies qx^2 = -py^2$
From $(i)$,$x^2 = 1 - py^2$. Substituting this into the condition:
$q(1 - py^2) = -py^2$
$q - qpy^2 = -py^2$
$q = y^2(qp - p) \implies y^2 = \frac{q}{p(q-1)}$
Similarly,$x^2 = \frac{p(1-q)}{q-p}$. Substituting $x^2$ and $y^2$ into $q x^2 + y^2 = 1$:
$q(\frac{p(1-q)}{q-p}) + \frac{q}{p(q-1)} = 1$
After simplifying,we get $p+q = 2pq$,which implies $\frac{1}{p} + \frac{1}{q} = 2$.

(C) Given curve is $y^4=a x^3$.
On differentiating with respect to $x$,we get:
$4 y^3 \frac{d y}{d x} = 3 a x^2$.
At the point $(a, a)$,the slope of the tangent is:
$\left(\frac{d y}{d x}\right)_{(a, a)} = \frac{3 a(a)^2}{4(a)^3} = \frac{3 a^3}{4 a^3} = \frac{3}{4}$.
The slope of the normal is the negative reciprocal of the tangent slope:
$m_{\text{normal}} = -\frac{1}{3/4} = -\frac{4}{3}$.
The equation of the normal at point $(a, a)$ is given by $y - y_1 = m(x - x_1)$:
$y - a = -\frac{4}{3}(x - a)$.
Multiplying by $3$: $3y - 3a = -4x + 4a$.
Rearranging the terms: $4x + 3y = 7a$.

(D) Given curves are $y^2=4x+4$ $(i)$ and $y^2=36(9-x)$ (ii).
To find the intersection points,equate the two expressions for $y^2$:
$4x+4 = 324-36x$
$40x = 320 \Rightarrow x=8$.
Substituting $x=8$ into $(i)$,$y^2 = 4(8)+4 = 36 \Rightarrow y = \pm 6$.
So,the intersection points are $(8,6)$ and $(8,-6)$.
Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$.
Differentiating (ii) with respect to $x$: $2y \frac{dy}{dx} = -36 \Rightarrow \frac{dy}{dx} = \frac{-18}{y}$.
At point $(8,6)$:
$m_1 = \frac{2}{6} = \frac{1}{3}$ and $m_2 = \frac{-18}{6} = -3$.
Since $m_1 \times m_2 = \frac{1}{3} \times (-3) = -1$,the tangents are perpendicular.
Thus,the angle between the curves is $90^{\circ}$ or $\frac{\pi}{2}$.

(B) Given the curve equation: $2y^4 = x^5$.
On differentiating both sides with respect to $x$,we get:
$8y^3 \frac{dy}{dx} = 5x^4$.
Thus,the slope of the tangent at $(2, 2)$ is:
$\left(\frac{dy}{dx}\right)_{(2, 2)} = \frac{5(2)^4}{8(2)^3} = \frac{5 \times 16}{8 \times 8} = \frac{80}{64} = \frac{5}{4}$.
The formula for the length of the subtangent is given by $\left| \frac{y}{dy/dx} \right|$.
Substituting the values,we get:
Length of subtangent $= \frac{2}{5/4} = 2 \times \frac{4}{5} = \frac{8}{5}$.

(D) The equations of the curves are $xy=2$ $\dots(i)$ and $x^2+4y=0$ $\dots(ii)$.
To find the point of intersection,substitute $y = -x^2/4$ from $(ii)$ into $(i)$:
$x(-x^2/4) = 2 \Rightarrow -x^3 = 8 \Rightarrow x = -2$.
Substituting $x = -2$ into $(ii)$,we get $4 + 4y = 0 \Rightarrow y = -1$.
So,the point of intersection is $(-2, -1)$.
For curve $(i)$,$y = 2/x$,so $dy/dx = -2/x^2$. At $x = -2$,$m_1 = -2/(-2)^2 = -2/4 = -1/2$.
For curve $(ii)$,$x^2 + 4y = 0$,so $2x + 4(dy/dx) = 0 \Rightarrow dy/dx = -x/2$. At $x = -2$,$m_2 = -(-2)/2 = 1$.
The angle $\theta$ between the curves is given by $\tan \theta = |(m_1 - m_2) / (1 + m_1 m_2)|$.
$\tan \theta = |(-1/2 - 1) / (1 + (-1/2)(1))| = |(-3/2) / (1/2)| = |-3| = 3$.

(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).
For curve $(i)$,differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.
For curve (ii),differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
To find the point of intersection,substitute $x=y^2$ into $xy=a^3$: $y^2 \cdot y = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.
Then $x = a^2$. So the point of intersection is $(a^2, a)$.
Let $m_1$ and $m_2$ be the slopes of the tangents at $(a^2, a)$.
$m_1 = \left(\frac{1}{2y}\right)_{(a^2, a)} = \frac{1}{2a}$.
$m_2 = \left(-\frac{y}{x}\right)_{(a^2, a)} = -\frac{a}{a^2} = -\frac{1}{a}$.
Since the curves cut orthogonally,$m_1 m_2 = -1$.
$\left(\frac{1}{2a}\right) \left(-\frac{1}{a}\right) = -1$.
$-\frac{1}{2a^2} = -1$.
$2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

(C) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y=x^2+3x+4$,we have $k=h^2+3h+4$ ... $(i)$.
The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = 2x+3$. At $(h, k)$,the slope is $m = 2h+3$.
The equation of the tangent at $(h, k)$ is $y-k = (2h+3)(x-h)$.
Since the tangent passes through $(0, 0)$,we substitute $x=0$ and $y=0$:
$0-k = (2h+3)(0-h) \Rightarrow -k = -2h^2-3h \Rightarrow k = 2h^2+3h$ ... $(ii)$.
Equating $(i)$ and $(ii)$:
$h^2+3h+4 = 2h^2+3h$
$h^2 = 4 \Rightarrow h = \pm 2$.
If $h = -2$,$k = (-2)^2+3(-2)+4 = 4-6+4 = 2$. So,$(\alpha, \beta) = (-2, 2)$.
If $h = 2$,$k = (2)^2+3(2)+4 = 4+6+4 = 14$. So,$(\gamma, \delta) = (2, 14)$.
Thus,$\beta+\delta = 2+14 = 16$.

(A) Given the curve $3y^2 = (x+5)^3$.
Differentiating with respect to $x$,we get $6y \frac{dy}{dx} = 3(x+5)^2$,which implies $\frac{dy}{dx} = \frac{(x+5)^2}{2y}$.
The length of the subtangent $ST = |\frac{y}{dy/dx}| = |\frac{y}{(x+5)^2 / 2y}| = |\frac{2y^2}{(x+5)^2}|$.
Substituting $y^2 = \frac{(x+5)^3}{3}$,we get $ST = |\frac{2(x+5)^3}{3(x+5)^2}| = |\frac{2(x+5)}{3}|$.
Thus,$(ST)^2 = \frac{4(x+5)^2}{9}$,so $9(ST)^2 = 4(x+5)^2$.
The length of the subnormal $SN = |y \frac{dy}{dx}| = |y \cdot \frac{(x+5)^2}{2y}| = |\frac{(x+5)^2}{2}|$.
Therefore,$8 SN = 8 \cdot \frac{(x+5)^2}{2} = 4(x+5)^2$.
Comparing the two results,we find $9(ST)^2 = 8 SN$.

(C) Given the curve $y = \int_0^x \frac{1}{1+t^3} dt$.
By applying the Leibniz integral rule,we differentiate the function with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \int_0^x \frac{1}{1+t^3} dt \right) = \frac{1}{1+x^3}$.
The slope of the tangent at $x = 1$ is given by the value of the derivative at that point:
$\left( \frac{dy}{dx} \right)_{x=1} = \frac{1}{1+(1)^3} = \frac{1}{1+1} = \frac{1}{2}$.

(A) Given the curve $y=4x^4+x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 16x^3+1$.
The slope of the tangent at $(0,0)$ is $m_1 = 16(0)^3+1 = 1$.
Let the point $P$ be $(a, b)$. The slope of the tangent at $P$ is $m_2 = 16a^3+1$.
Since the tangents are perpendicular,$m_1 \times m_2 = -1$.
$1 \times (16a^3+1) = -1$.
$16a^3 = -2$ $\Rightarrow a^3 = -\frac{1}{8}$ $\Rightarrow a = -\frac{1}{2}$.
Since $P$ lies on the curve,$b = 4(-\frac{1}{2})^4 + (-\frac{1}{2}) = 4(\frac{1}{16}) - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.
Thus,the point $P$ is $\left(-\frac{1}{2}, -\frac{1}{4}\right)$.

(D) Given the curve $y^2=4x$ and the point $P(1,2)$.
The slope of the tangent at $(1,2)$ is found by differentiating $y^2=4x$: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$. At $(1,2)$,the slope $m_t = \frac{2}{2} = 1$.
The equation of the tangent is $y-2 = 1(x-1) \Rightarrow y = x+1$.
The tangent intersects the $Y$-axis $(x=0)$ at $A(0,1)$.
The slope of the normal $m_n = -\frac{1}{m_t} = -1$.
The equation of the normal is $y-2 = -1(x-1) \Rightarrow y = -x+3$.
The normal intersects the $Y$-axis $(x=0)$ at $B(0,3)$.
The triangle is formed by vertices $P(1,2)$,$A(0,1)$,and $B(0,3)$.
The base of the triangle along the $Y$-axis is the distance between $A(0,1)$ and $B(0,3)$,which is $|3-1| = 2$ units.
The height of the triangle is the perpendicular distance from $P(1,2)$ to the $Y$-axis,which is the $x$-coordinate of $P$,i.e.,$1$ unit.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$ square unit.

(B) Given the parametric equations of the curve are $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$.
To find the slope of the tangent,we differentiate $x$ and $y$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(1+\cos \theta)$ and $\frac{dy}{d\theta} = a\sin \theta$.
Thus,the slope of the tangent $\frac{dy}{dx}$ is given by:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin \theta}{a(1+\cos \theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2)$.
Given that the tangent is inclined at an angle $\frac{\pi}{4}$ to the $x$-axis,the slope is $\tan(\frac{\pi}{4}) = 1$.
Equating the slopes: $\tan(\theta/2) = \tan(\frac{\pi}{4})$,which implies $\theta/2 = \frac{\pi}{4}$,so $\theta = \frac{\pi}{2}$.
Substituting $\theta = \frac{\pi}{2}$ into the expressions for $x$ and $y$:
$x = a(\frac{\pi}{2} + \sin(\frac{\pi}{2})) = a(\frac{\pi}{2} + 1)$.
$y = a(1 - \cos(\frac{\pi}{2})) = a(1 - 0) = a$.
Therefore,the coordinates of point $P$ are $\left[a\left(\frac{\pi}{2}+1\right), a\right]$.

(A) To find the point of intersection,set the equations equal: $3x^2-2x-1 = x^3-1$.
$x^3-3x^2+2x = 0$.
$x(x^2-3x+2) = 0$.
$x(x-1)(x-2) = 0$.
So,$x=0, 1, 2$.
For $x=0$,$y=-1$ (not in the first quadrant).
For $x=1$,$y=0$ (on the axis,not strictly in the first quadrant).
For $x=2$,$y=3(2)^2-2(2)-1 = 12-4-1 = 7$.
The point of intersection in the first quadrant is $(2, 7)$.
Now,find the slopes of the tangents at $(2, 7)$:
For $y=3x^2-2x-1$,$dy/dx = 6x-2$. At $x=2$,$m_1 = 6(2)-2 = 10$.
For $y=x^3-1$,$dy/dx = 3x^2$. At $x=2$,$m_2 = 3(2)^2 = 12$.
The angle $\theta$ between the curves is given by $\tan \theta = |(m_1-m_2)/(1+m_1m_2)|$.
$\tan \theta = |(10-12)/(1+10 \times 12)| = |-2/121| = 2/121$.
Thus,$\theta = \operatorname{Tan}^{-1}(2/121)$.

(C) Given point $P(5,2)$ and slope of the tangent $m_t = \frac{7}{2}$.
Equation of the tangent at $P(5,2)$ is $y - 2 = \frac{7}{2}(x - 5) \implies 2y - 4 = 7x - 35 \implies 7x - 2y = 31$.
The $x$-intercept of the tangent is found by setting $y=0$: $7x = 31 \implies x = \frac{31}{7}$. So,the tangent meets the $x$-axis at $A(\frac{31}{7}, 0)$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{2}{7}$.
Equation of the normal at $P(5,2)$ is $y - 2 = -\frac{2}{7}(x - 5) \implies 7y - 14 = -2x + 10 \implies 2x + 7y = 24$.
The $x$-intercept of the normal is found by setting $y=0$: $2x = 24 \implies x = 12$. So,the normal meets the $x$-axis at $B(12, 0)$.
The triangle is formed by points $P(5,2)$,$A(\frac{31}{7}, 0)$,and $B(12, 0)$.
The base of the triangle on the $x$-axis is $|12 - \frac{31}{7}| = |\frac{84 - 31}{7}| = \frac{53}{7}$.
The height of the triangle is the $y$-coordinate of $P$,which is $2$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{53}{7} \times 2 = \frac{53}{7}$ sq. units.

(A) Let the point $P$ be $(x_1, y_1)$. The curve is $y^2 = x^3 - x + 1$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 3x^2 - 1$,so $\frac{dy}{dx} = \frac{3x_1^2 - 1}{2y_1}$.
The slope of the normal at $P$ is $m_n = -\frac{1}{dy/dx} = -\frac{2y_1}{3x_1^2 - 1}$.
Since the normal makes equal intercepts on the axes,its slope must be $\pm 1$. Given the curve,we test $m_n = -1$ (as the normal equation $x+y=c$ or $x-y=c$ implies slope $\pm 1$).
If $m_n = -1$,then $\frac{2y_1}{3x_1^2 - 1} = 1 \implies 2y_1 = 3x_1^2 - 1$.
Substituting $y_1^2 = x_1^3 - x_1 + 1$ and $y_1 = \frac{3x_1^2 - 1}{2}$,we get $(\frac{3x_1^2 - 1}{2})^2 = x_1^3 - x_1 + 1$. Solving this,we find $x_1 = 1$,which gives $y_1 = 1$. Thus $P = (1, 1)$.
The slope of the tangent at $(1, 1)$ is $m_t = \frac{3(1)^2 - 1}{2(1)} = \frac{2}{2} = 1$.
The equation of the tangent is $y - 1 = 1(x - 1)$,which simplifies to $y = x$,or $x - y = 0$.

(D) Given the curve $xy^2 + x^2y = 12$. Differentiating with respect to $x$:
$y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$.
At the point $(1, 3)$,we have $3^2 + 2(1)(3) \frac{dy}{dx} + 2(1)(3) + (1)^2 \frac{dy}{dx} = 0$.
$9 + 6 \frac{dy}{dx} + 6 + \frac{dy}{dx} = 0 \implies 7 \frac{dy}{dx} = -15 \implies \frac{dy}{dx} = -\frac{15}{7}$.
The equation of the tangent at $(1, 3)$ is $y - 3 = -\frac{15}{7}(x - 1)$.
For $T$,set $y = 0$: $-3 = -\frac{15}{7}(x - 1) \implies 21 = 15(x - 1) \implies x - 1 = \frac{21}{15} = \frac{7}{5} \implies x = 1 + \frac{7}{5} = \frac{12}{5}$. So $T = (\frac{12}{5}, 0)$.
The equation of the normal at $(1, 3)$ is $y - 3 = \frac{7}{15}(x - 1)$.
For $N$,set $y = 0$: $-3 = \frac{7}{15}(x - 1) \implies -45 = 7(x - 1) \implies x - 1 = -\frac{45}{7} \implies x = 1 - \frac{45}{7} = -\frac{38}{7}$. So $N = (-\frac{38}{7}, 0)$.
$TN = |\frac{12}{5} - (-\frac{38}{7})| = |\frac{12}{5} + \frac{38}{7}| = |\frac{84 + 190}{35}| = \frac{274}{35}$.

(C) The slope of the line $y = x - 3$ is $m = 1$.
Since $P(x_1, y_1)$ is the closest point on the curve $y = x^2 - x + 1$ to the line,the tangent at $P$ must be parallel to the given line.
Thus,the derivative $\frac{dy}{dx} = 2x - 1$ must equal $1$.
$2x_1 - 1 = 1 \implies 2x_1 = 2 \implies x_1 = 1$.
Substituting $x_1 = 1$ into the curve equation: $y_1 = (1)^2 - 1 + 1 = 1$.
So,the point $P$ is $(1, 1)$.
The perpendicular distance from $P(1, 1)$ to the line $3x + 4y - 2 = 0$ is given by the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
$d = \frac{|3(1) + 4(1) - 2|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 4 - 2|}{\sqrt{9 + 16}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$.

(B) Given curves are $y^2 = 12x - 3$ and $y^2 = 12 - kx$.
Let the point of intersection be $(x_1, y_1)$.
For $y^2 = 12x - 3$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 12 \implies \frac{dy}{dx} = \frac{6}{y}$. Let $m_1 = \frac{6}{y_1}$.
For $y^2 = 12 - kx$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = -k \implies \frac{dy}{dx} = -\frac{k}{2y}$. Let $m_2 = -\frac{k}{2y_1}$.
Since the curves cut orthogonally,$m_1 m_2 = -1 \implies (\frac{6}{y_1})(-\frac{k}{2y_1}) = -1 \implies \frac{3k}{y_1^2} = 1 \implies y_1^2 = 3k$.
At intersection point $(x_1, y_1)$,$12x_1 - 3 = 12 - kx_1 \implies x_1(12 + k) = 15 \implies x_1 = \frac{15}{12+k}$.
Also $y_1^2 = 12 - kx_1 = 3k \implies 12 - k(\frac{15}{12+k}) = 3k \implies 12(12+k) - 15k = 3k(12+k) \implies 144 + 12k - 15k = 36k + 3k^2 \implies 3k^2 + 39k - 144 = 0 \implies k^2 + 13k - 48 = 0 \implies (k+16)(k-3) = 0$.
Since $k > 0$,$k = 3$.
The curve is $y^2 = 12 - 3x$. At $x = 1$,$y^2 = 12 - 3(1) = 9 \implies y = 3$ (taking $b=3$).
The slope at $(1, 3)$ is $m = -\frac{3}{2(3)} = -\frac{1}{2}$.
The length of the sub-tangent is $|\frac{y}{dy/dx}| = |\frac{3}{-1/2}| = 6$.

(C) Given the curve $\frac{x^n}{a^n} + \frac{y^n}{b^n} = 2$.
At the point $(a, b)$,we check if it lies on the curve: $\frac{a^n}{a^n} + \frac{b^n}{b^n} = 1 + 1 = 2$. Thus,$(a, b)$ is on the curve.
Differentiating the equation with respect to $x$: $\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \frac{dy}{dx} = 0$.
So,$\frac{dy}{dx} = -\frac{x^{n-1}}{a^n} \cdot \frac{b^n}{y^{n-1}} = -\frac{b^n x^{n-1}}{a^n y^{n-1}}$.
At the point $(a, b)$,the slope of the tangent $m = -\frac{b^n a^{n-1}}{a^n b^{n-1}} = -\frac{b}{a}$.
The equation of the tangent line at $(a, b)$ is $y - b = -\frac{b}{a}(x - a)$,which simplifies to $ay - ab = -bx + ab$,or $bx + ay = 2ab$.
Dividing by $ab$,we get $\frac{x}{a} + \frac{y}{b} = 2$.
Since this matches the given line,the line is a tangent for all values of $n > 1$.

(B) Given the curve $y=x^3-2x^2+3x-4$ and the line $y=-2$.
At the point of intersection $P(h, k)$,we have $x^3-2x^2+3x-4 = -2$.
$\Rightarrow x^3-2x^2+3x-2 = 0$.
By testing values,for $x=1$,$1-2+3-2 = 0$. Thus,the point $P$ is $(1, -2)$.
Now,find the slope of the tangent at $P(1, -2)$:
$\frac{dy}{dx} = 3x^2-4x+3$.
At $x=1$,$\frac{dy}{dx} = 3(1)^2-4(1)+3 = 3-4+3 = 2$.
The equation of the tangent at $(1, -2)$ is $y - (-2) = 2(x - 1)$.
$\Rightarrow y+2 = 2x-2
\Rightarrow y = 2x-4$.
This tangent meets the $X$-axis where $y=0$:
$0 = 2x_1 - 4
\Rightarrow 2x_1 = 4
\Rightarrow x_1 = 2$.

(D) Given the curve $y^3=4 x^5$.
Differentiating both sides with respect to $x$,we get $3 y^2 \frac{d y}{d x} = 20 x^4$.
Thus,$\frac{d y}{d x} = \frac{20 x^4}{3 y^2}$.
At the point $(4, 16)$,the slope of the tangent is $m_t = \frac{20(4)^4}{3(16)^2} = \frac{20 \times 256}{3 \times 256} = \frac{20}{3}$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{3}{20}$.
The equation of the normal at $(4, 16)$ is $y - 16 = -\frac{3}{20}(x - 4)$.
Multiplying by $20$,we get $20y - 320 = -3x + 12$.
Rearranging the terms,we get $3x + 20y = 332$.

(C) Let $m_1$ be the slope of the tangent to the curve $y=f(x)$ at point $P$. Given $m_1 = \frac{dy}{dx} = Q(x)$.
Let $m_2$ be the slope of the tangent to the curve $x=g(y)$ at point $P$. Given $\frac{dx}{dy} = -Q(x)$,we know that $m_2 = \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{-Q(x)}$.
Since $m_1 \times m_2 = Q(x) \times \left(-\frac{1}{Q(x)}\right) = -1$,the product of the slopes of the tangents at the point of intersection is $-1$.
Therefore,the tangent drawn at $P$ to one curve is normal to the other curve at $P$.

(B) Given curve: $y^3 - 3xy + 2 = 0$.
Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$.
$\frac{dy}{dx} (3y^2 - 3x) = 3y \implies \frac{dy}{dx} = \frac{y}{y^2 - x}$.
For a horizontal tangent,$\frac{dy}{dx} = 0 \implies y = 0$.
Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \implies 2 = 0$,which is impossible. Thus,$n(A) = 0$.
For a vertical tangent,$\frac{dy}{dx} = \infty \implies y^2 - x = 0 \implies x = y^2$.
Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0 \implies y^3 - 3y^3 + 2 = 0 \implies -2y^3 = -2 \implies y^3 = 1 \implies y = 1$.
If $y = 1$,then $x = 1^2 = 1$. The point is $(1, 1)$. Thus,$n(B) = 1$.
Therefore,$n(A) + n(B) = 0 + 1 = 1$.

(B) Given the curve $x = 2(\cos 2t + t \sin 2t)$ and $y = 4(\sin 2t - t \cos 2t)$.
First,find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:
$\frac{dx}{dt} = 2(-2 \sin 2t + \sin 2t + 2t \cos 2t) = 2(2t \cos 2t - \sin 2t)$.
$\frac{dy}{dt} = 4(2 \cos 2t - \cos 2t + 2t \sin 2t) = 4(\cos 2t + 2t \sin 2t)$.
At $t = \frac{\pi}{4}$:
$\frac{dx}{dt} = 2(2(\frac{\pi}{4}) \cos \frac{\pi}{2} - \sin \frac{\pi}{2}) = 2(0 - 1) = -2$.
$\frac{dy}{dt} = 4(\cos \frac{\pi}{2} + 2(\frac{\pi}{4}) \sin \frac{\pi}{2}) = 4(0 + \frac{\pi}{2}) = 2\pi$.
Therefore,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\pi}{-2} = -\pi$.
The slope of the normal is $-\frac{1}{dy/dx} = \frac{1}{\pi}$.
At $t = \frac{\pi}{4}$,$y = 4(\sin \frac{\pi}{2} - \frac{\pi}{4} \cos \frac{\pi}{2}) = 4(1 - 0) = 4$.
The length of the normal is given by $|y| \sqrt{1 + (\frac{dy}{dx})^2} = |4| \sqrt{1 + (-\pi)^2} = 4 \sqrt{1 + \pi^2}$.

(C) Given the curve $y = \cos(x+y)$.
Taking the derivative with respect to $x$:
$\frac{dy}{dx} = -\sin(x+y) \cdot (1 + \frac{dy}{dx})$
$\frac{dy}{dx} (1 + \sin(x+y)) = -\sin(x+y)$
$\frac{dy}{dx} = \frac{-\sin(x+y)}{1 + \sin(x+y)}$
The equation of the tangent is $x + 2y = k$,which can be written as $y = -\frac{1}{2}x + \frac{k}{2}$.
The slope of the tangent is $m = -\frac{1}{2}$.
Equating the slopes: $\frac{-\sin(x+y)}{1 + \sin(x+y)} = -\frac{1}{2}$
$2\sin(x+y) = 1 + \sin(x+y)$
$\sin(x+y) = 1$
Since $\sin(x+y) = 1$,we have $y = \cos(x+y) = \cos(\frac{\pi}{2} + 2n\pi) = 0$.
Substituting $y = 0$ into $\sin(x+y) = 1$,we get $\sin(x) = 1$,so $x = \frac{\pi}{2}$.
Now,substitute $x = \frac{\pi}{2}$ and $y = 0$ into the tangent equation $x + 2y = k$:
$\frac{\pi}{2} + 2(0) = k$
$k = \frac{\pi}{2}$.

(C) Given the curve $y = x^3 - 10x^2 + 31x - 30$.
First,find the derivative: $\frac{dy}{dx} = 3x^2 - 20x + 31$.
The slope of the normal is given by $-\frac{1}{\frac{dy}{dx}}$.
Setting this equal to $-\frac{1}{14}$:
$-\frac{1}{3x^2 - 20x + 31} = -\frac{1}{14} \implies 3x^2 - 20x + 31 = 14$.
$3x^2 - 20x + 17 = 0$.
Factoring the quadratic: $(3x - 17)(x - 1) = 0$.
This gives $x = 1$ or $x = \frac{17}{3}$.
Since the problem states $x$ is an integer,we take $x = 1$.
Substituting $x = 1$ into the curve equation: $y = (1)^3 - 10(1)^2 + 31(1) - 30 = 1 - 10 + 31 - 30 = -8$.
So,the point $P$ is $(1, -8)$.
The slope of the tangent at $x = 1$ is $\frac{dy}{dx} = 3(1)^2 - 20(1) + 31 = 14$.
The equation of the tangent line is $y - (-8) = 14(x - 1) \implies y + 8 = 14x - 14 \implies y = 14x - 22$.
For the $x$-intercept,set $y = 0$: $0 = 14x - 22 \implies 14x = 22 \implies x = \frac{22}{14} = \frac{11}{7}$.

(B) Given $\frac{dy}{dx} = 3x^2 - 5$.
Integrating both sides,we get $y = \int (3x^2 - 5) dx = x^3 - 5x + C$.
Since $f(1) = 2$,we have $2 = 1^3 - 5(1) + C$,which gives $2 = 1 - 5 + C$,so $C = 6$.
Thus,the equation of the curve is $y = x^3 - 5x + 6$.
The slope of the tangent at $(1, 2)$ is $\left. \frac{dy}{dx} \right|_{(1, 2)} = 3(1)^2 - 5 = -2$.
The equation of the tangent at $(1, 2)$ is $y - 2 = -2(x - 1)$,which simplifies to $y = -2x + 4$.
To find the intersection point,set the curve equation equal to the tangent equation: $x^3 - 5x + 6 = -2x + 4$.
This simplifies to $x^3 - 3x + 2 = 0$.
Factoring the cubic equation,we get $(x - 1)^2(x + 2) = 0$.
The roots are $x = 1$ and $x = -2$.
For $x = 1$,$y = 2$ (the point of tangency).
For $x = -2$,$y = -2(-2) + 4 = 8$.
Therefore,the tangent intersects the curve at the point $(-2, 8)$.

(D) Given curves are $x^2 y - x^3 = 8$ $(1)$ and $y^3 - x y^2 = 32$ $(2)$.
Differentiating $(1)$ with respect to $x$: $x^2 \frac{dy}{dx} + 2xy - 3x^2 = 0 \Rightarrow \frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2} = 3 - 2(\frac{y}{x}) = m_1$.
Differentiating $(2)$ with respect to $x$: $3y^2 \frac{dy}{dx} - (x \cdot 2y \frac{dy}{dx} + y^2) = 0 \Rightarrow \frac{dy}{dx}(3y^2 - 2xy) = y^2 \Rightarrow \frac{dy}{dx} = \frac{y^2}{3y^2 - 2xy} = \frac{y}{3y - 2x} = m_2$.
Dividing $(2)$ by $(1)$: $\frac{y^2(y - x)}{x^2(y - x)} = \frac{32}{8} = 4 \Rightarrow \frac{y^2}{x^2} = 4 \Rightarrow y = 2x$ (since $h, k \in N$).
Substituting $y = 2x$ into $(1)$: $x^2(2x) - x^3 = 8 \Rightarrow x^3 = 8 \Rightarrow x = 2$. Thus,$y = 4$.
At $P(2, 4)$,$m_1 = 3 - 2(\frac{4}{2}) = 3 - 4 = -1$.
At $P(2, 4)$,$m_2 = \frac{4}{3(4) - 2(2)} = \frac{4}{12 - 4} = \frac{4}{8} = \frac{1}{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{1/2 - (-1)}{1 + (-1)(1/2)}| = |\frac{3/2}{1/2}| = 3$.

(D) Given the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$.
Thus,$\frac{dy}{dx} = \frac{2t - 4}{2t - 7}$.
At the point $(1, 2)$,we have $t^2 - 7t + 7 = 1 \implies t^2 - 7t + 6 = 0 \implies (t-1)(t-6) = 0$,so $t = 1$ or $t = 6$.
Also,$t^2 - 4t - 10 = 2 \implies t^2 - 4t - 12 = 0 \implies (t-6)(t+2) = 0$,so $t = 6$ or $t = -2$.
The common value is $t = 6$.
The slope of the tangent at $t = 6$ is $\frac{2(6) - 4}{2(6) - 7} = \frac{8}{5}$.
The slope of the normal $m$ is $-\frac{1}{8/5} = -\frac{5}{8}$.
The equation of the normal at $(1, 2)$ is $y - 2 = -\frac{5}{8}(x - 1)$.
$8y - 16 = -5x + 5 \implies 5x + 8y - 21 = 0$.
Here $a = 5, b = 8, c = -21$. The $G$.$C$.$D$. of $(5, 8, -21)$ is $1$.
Then $a + b + c = 5 + 8 - 21 = -8$.
Finally,$m(a + b + c) = (-\frac{5}{8})(-8) = 5$.

(D) Given the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$.
The slope of the tangent is given by the derivative $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$.
Since the slope of the tangent is $2$,we set $\frac{dy}{dx} = 2$:
$4x^3 - 18x^2 + 26x - 10 = 2$
$4x^3 - 18x^2 + 26x - 12 = 0$
Dividing by $2$,we get $2x^3 - 9x^2 + 13x - 6 = 0$.
By testing integer roots,we find $(x-1)$ and $(x-2)$ are factors. Factoring gives $(x-1)(x-2)(2x-3) = 0$.
The roots are $x = 1, x = 2, x = 1.5$.
Since $x_1, x_2 \in \mathbb{N}$,we choose $x_1 = 1$ and $x_2 = 2$.
For $x_1 = 1$,$y_1 = 1^4 - 6(1)^3 + 13(1)^2 - 10(1) + 5 = 1 - 6 + 13 - 10 + 5 = 3$.
For $x_2 = 2$,$y_2 = 2^4 - 6(2)^3 + 13(2)^2 - 10(2) + 5 = 16 - 48 + 52 - 20 + 5 = 5$.
Thus,$x_1x_2 - y_1y_2 = (1 \times 2) - (3 \times 5) = 2 - 15 = -13$.

(A) Given the equation of the curve is $\sin y = \sqrt{3} x \sin \left(\frac{\pi}{6} + y\right) \quad (i)$.
At $x = 0$,$\sin y = 0$,which implies $y = 0$.
Now,differentiating equation $(i)$ with respect to $x$ using the product rule:
$\cos y \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + y\right) + \sqrt{3} x \cos \left(\frac{\pi}{6} + y\right) \frac{dy}{dx}$.
At the point $(0, 0)$,substituting $x = 0$ and $y = 0$:
$\cos(0) \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6} + 0\right) + \sqrt{3}(0) \cos \left(\frac{\pi}{6} + 0\right) \frac{dy}{dx}$.
$1 \cdot \frac{dy}{dx} = \sqrt{3} \sin \left(\frac{\pi}{6}\right) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$.
Thus,the slope of the tangent $m_t = \frac{\sqrt{3}}{2}$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$.
The equation of the normal at $(0, 0)$ is given by $y - y_1 = m_n(x - x_1)$:
$y - 0 = -\frac{2}{\sqrt{3}}(x - 0)$.
$\sqrt{3}y = -2x$,which simplifies to $2x + \sqrt{3}y = 0$.

(B) The given curve is $y=x^2$. When it is shifted by '$a$' units along the positive $X$-axis,the new curve becomes $y=(x-a)^2$.
To find the intersection point,we set the equations equal:
$x^2 = (x-a)^2$
$x^2 = x^2 - 2ax + a^2$
$2ax = a^2$
Since $a \neq 0$,we get $x = \frac{a}{2}$.
Substituting $x = \frac{a}{2}$ into $y=x^2$,we get $y = \frac{a^2}{4}$.
The intersection point is $(\frac{a}{2}, \frac{a^2}{4})$.
Now,find the slopes of the tangents at this point:
For $y=x^2$,$\frac{dy}{dx} = 2x$. At $x=\frac{a}{2}$,$m_1 = 2(\frac{a}{2}) = a$.
For $y=(x-a)^2$,$\frac{dy}{dx} = 2(x-a)$. At $x=\frac{a}{2}$,$m_2 = 2(\frac{a}{2}-a) = -a$.
The angle $\theta$ between the curves is given by:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{a - (-a)}{1 + (a)(-a)} \right| = \left| \frac{2a}{1 - a^2} \right| = \frac{2|a|}{|1 - a^2|}$.
Thus,option $(b)$ is correct.

(B) Given curves are $x^2-y^2=4$ and $y^2=3x$.
Substituting $y^2=3x$ into the first equation: $x^2-3x-4=0$.
$(x-4)(x+1)=0$. Since $y^2=3x$,$x$ must be positive,so $x=4$.
Then $y^2=12$,so $y=\pm 2\sqrt{3}$. Let the point of intersection be $(4, 2\sqrt{3})$.
For $x^2-y^2=4$,differentiating w.r.t $x$: $2x-2y \frac{dy}{dx}=0 \implies \frac{dy}{dx}=\frac{x}{y}$.
At $(4, 2\sqrt{3})$,$m_1 = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.
For $y^2=3x$,differentiating w.r.t $x$: $2y \frac{dy}{dx}=3 \implies \frac{dy}{dx}=\frac{3}{2y}$.
At $(4, 2\sqrt{3})$,$m_2 = \frac{3}{2(2\sqrt{3})} = \frac{3}{4\sqrt{3}} = \frac{\sqrt{3}}{4}$.
$\tan \theta = \left| \frac{m_1-m_2}{1+m_1 m_2} \right| = \left| \frac{\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{4}}{1+(\frac{2}{\sqrt{3}})(\frac{\sqrt{3}}{4})} \right| = \left| \frac{\frac{8-3}{4\sqrt{3}}}{1+\frac{1}{2}} \right| = \left| \frac{5}{4\sqrt{3}} \times \frac{2}{3} \right| = \frac{5}{6\sqrt{3}}$.