Tangent and Normal Questions in English

(D) The curves are $y_1 = 3^x$ and $y_2 = 7^x$.
They intersect where $3^x = 7^x$,which implies $x=0$.
At $x=0$,$y=3^0=1$. So the point of intersection is $(0, 1)$.
The slopes of the tangents are $m_1 = \frac{dy_1}{dx} = 3^x \ln 3$ and $m_2 = \frac{dy_2}{dx} = 7^x \ln 7$.
At $(0, 1)$,$m_1 = \ln 3$ and $m_2 = \ln 7$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
$\tan \theta = |\frac{\ln 7 - \ln 3}{1 + (\ln 3)(\ln 7)}| = \frac{\ln(7/3)}{1 + (\ln 3)(\ln 7)}$.
Since $\log x$ usually denotes $\log_{10} x$ and $\ln x$ denotes $\log_e x$,the expression is equivalent to $\frac{\log(7/3)}{1 + (\log 3)(\log 7)}$ if the base is consistent.

(C) Given the curve $x^2+2xy-3y^2=0$.
Differentiating with respect to $x$: $2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0$.
At point $(2,2)$: $2(2) + 2(2) + 2(2)\frac{dy}{dx} - 6(2)\frac{dy}{dx} = 0$.
$4 + 4 + 4\frac{dy}{dx} - 12\frac{dy}{dx} = 0 \implies 8 - 8\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = 1$.
The slope of the tangent at $(2,2)$ is $m = 1$.
The slope of the normal is $m' = -\frac{1}{m} = -1$.
The equation of the normal at $(2,2)$ is $y - 2 = -1(x - 2) \implies y - 2 = -x + 2 \implies x + y - 4 = 0$.
The length of the perpendicular from the origin $(0,0)$ to the line $x + y - 4 = 0$ is given by $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
$d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

(A) Given the curve equation: $(1+x^2)y = 2-x$.
To find the point where the curve crosses the $X$-axis,set $y=0$:
$(1+x^2)(0) = 2-x \implies 2-x = 0 \implies x=2$.
So,the point of contact is $(2, 0)$.
Now,differentiate the equation with respect to $x$:
$(1+x^2) \frac{dy}{dx} + y(2x) = -1$.
Substitute $x=2$ and $y=0$ to find the slope $m$:
$(1+2^2) \frac{dy}{dx} + 0(2 \times 2) = -1
(5) \frac{dy}{dx} = -1
\frac{dy}{dx} = -\frac{1}{5}$.
The equation of the tangent at $(2, 0)$ with slope $m = -\frac{1}{5}$ is:
$y - 0 = -\frac{1}{5}(x - 2)
5y = -x + 2
x + 5y = 2$.
Thus,the correct option is $A$.

(A) The given curve is $y = b e^{-x / a}$.
To find the point where the curve crosses the $Y$ axis,we set $x = 0$.
Substituting $x = 0$ into the equation,we get $y = b e^0 = b$.
So,the point of contact is $(0, b)$.
Now,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = b \cdot e^{-x / a} \cdot (-1 / a) = -\frac{b}{a} e^{-x / a}$.
At the point $(0, b)$,the slope $m$ is:
$m = \left. \frac{dy}{dx} \right|_{(0, b)} = -\frac{b}{a} e^0 = -\frac{b}{a}$.
The equation of the tangent line passing through $(x_1, y_1) = (0, b)$ with slope $m = -\frac{b}{a}$ is given by $y - y_1 = m(x - x_1)$.
$y - b = -\frac{b}{a}(x - 0)$.
$y - b = -\frac{b}{a}x$.
Dividing both sides by $b$,we get $\frac{y}{b} - 1 = -\frac{x}{a}$.
Rearranging the terms,we get $\frac{x}{a} + \frac{y}{b} = 1$.

(B) Given the curve $y^2 = \frac{x^3}{9}$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = \frac{3x^2}{9} = \frac{x^2}{3}$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.
The slope of the normal at point $(x_1, y_1)$ is $m_n = -\frac{1}{dy/dx} = -\frac{6y_1}{x_1^2}$.
The equation of the normal is $y - y_1 = -\frac{6y_1}{x_1^2}(x - x_1)$.
Since the normal makes equal intercepts with the axes,its slope must be $\pm 1$. Given the geometry,the slope is $-1$.
So,$-\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.
Substitute $y_1^2 = \frac{x_1^3}{9}$ into the equation: $y_1 = \frac{x_1^2}{6} \implies y_1^2 = \frac{x_1^4}{36}$.
Equating the two expressions for $y_1^2$: $\frac{x_1^3}{9} = \frac{x_1^4}{36} \implies 4x_1^3 = x_1^4 \implies x_1 = 4$ (since $x_1 \neq 0$).
If $x_1 = 4$,then $y_1^2 = \frac{64}{9} \implies y_1 = \pm \frac{8}{3}$.
Thus,the points are $(4, \pm \frac{8}{3})$.

(C) Given the curve $xy = a^2$. Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
At the point $(x_1, y_1)$,the slope of the tangent is $m = -\frac{y_1}{x_1}$.
The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{x_1}(x - x_1)$.
Multiplying by $x_1$,we get $x_1 y - x_1 y_1 = -y_1 x + x_1 y_1$,which simplifies to $x_1 y + y_1 x = 2x_1 y_1$.
Since $x_1 y_1 = a^2$,the equation becomes $x_1 y + y_1 x = 2a^2$.
Dividing by $2a^2$,we get $\frac{x}{2a^2/y_1} + \frac{y}{2a^2/x_1} = 1$.
This is the intercept form of a line $\frac{x}{A} + \frac{y}{B} = 1$,where the x-intercept $A = \frac{2a^2}{y_1}$ and the y-intercept $B = \frac{2a^2}{x_1}$.
The area of the triangle formed by the axes and the tangent is $\frac{1}{2} \times |A| \times |B| = \frac{1}{2} \times \frac{2a^2}{y_1} \times \frac{2a^2}{x_1} = \frac{2a^4}{x_1 y_1}$.
Since $x_1 y_1 = a^2$,the area is $\frac{2a^4}{a^2} = 2a^2$ sq. units.

(B) The given equation of the curve is $y(x-2)(x-3)=x+6$.
At the $Y$-axis,$x=0$. Substituting $x=0$ into the equation: $y(0-2)(0-3)=0+6 \Rightarrow y(-2)(-3)=6 \Rightarrow 6y=6 \Rightarrow y=1$.
So,the point of intersection is $(0, 1)$.
Now,rewrite the equation as $y = \frac{x+6}{x^2-5x+6}$.
Differentiating 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 $x=0$,the slope of the tangent $m_t = \frac{(0-0+6)(1) - (0+6)(0-5)}{(0-0+6)^2} = \frac{6 - (-30)}{36} = \frac{36}{36} = 1$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal at $(0, 1)$ is $y - 1 = -1(x - 0)$,which simplifies to $y = -x + 1$ or $x + y = 1$.
Checking the options,for $(\frac{1}{2}, \frac{1}{2})$,we have $\frac{1}{2} + \frac{1}{2} = 1$,which satisfies the equation.
Thus,the normal passes through $(\frac{1}{2}, \frac{1}{2})$.

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

(A) 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,the slope of the tangent is $\frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.
Let the point of tangency be $(x_1, y_1)$. 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 $\sqrt{x_1}y - \sqrt{x_1}y_1 = -\sqrt{y_1}x + \sqrt{y_1}x_1$.
Rearranging gives $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1}y_1 + \sqrt{y_1}x_1 = \sqrt{x_1y_1}(\sqrt{x_1}+\sqrt{y_1})$.
Since $\sqrt{x_1}+\sqrt{y_1}=\sqrt{a}$,the equation becomes $\sqrt{y_1}x + \sqrt{x_1}y = \sqrt{x_1y_1a}$.
Dividing by $\sqrt{x_1y_1a}$,we get $\frac{x}{\sqrt{x_1a}} + \frac{y}{\sqrt{y_1a}} = 1$.
The $x$-intercept is $\sqrt{x_1a}$ and the $y$-intercept is $\sqrt{y_1a}$.
The sum of intercepts is $\sqrt{a}(\sqrt{x_1}+\sqrt{y_1}) = \sqrt{a} \times \sqrt{a} = a$.

(B) Given equation of the curve is $y=1-e^{\frac{x}{3}} \dots (i)$.
Since the curve intersects the $Y$-axis,we set $x=0$.
Substituting $x=0$ in $(i)$,we get $y=1-e^{0}=1-1=0$.
Thus,the point of intersection is $(0, 0)$.
The slope of the tangent is given by $\frac{dy}{dx}$.
Differentiating $(i)$ with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{3} e^{\frac{x}{3}}$.
At the point $(0, 0)$,the slope $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{3} e^{0} = -\frac{1}{3}$.
The equation of the tangent line passing through $(0, 0)$ with slope $m = -\frac{1}{3}$ is $y - 0 = -\frac{1}{3}(x - 0)$.
Multiplying by $3$,we get $3y = -x$,which simplifies to $x + 3y = 0$.

(A) Given the curve $y(x) = 1 + \sqrt{4x-3}$.
First,we find the derivative to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{1}{2\sqrt{4x-3}} \times 4 = \frac{2}{\sqrt{4x-3}}$.
Given the slope of the tangent at point $P$ is $\frac{2}{3}$,we set $\frac{2}{\sqrt{4x-3}} = \frac{2}{3}$.
This implies $\sqrt{4x-3} = 3$,so $4x-3 = 9$,which gives $4x = 12$,or $x = 3$.
Substituting $x = 3$ into the curve equation,we get $y = 1 + \sqrt{4(3)-3} = 1 + \sqrt{9} = 1 + 3 = 4$.
Thus,the point $P$ is $(3, 4)$.
The slope of the normal at $P$ is the negative reciprocal of the tangent slope: $m_{normal} = -\frac{1}{2/3} = -\frac{3}{2}$.
The equation of the normal at $(3, 4)$ is $y - 4 = -\frac{3}{2}(x - 3)$.
Multiplying by $2$,we get $2y - 8 = -3x + 9$,which simplifies to $3x + 2y - 17 = 0$.
Checking the options:
For $(1, 7)$: $3(1) + 2(7) - 17 = 3 + 14 - 17 = 0$. This satisfies the equation.
Therefore,the normal passes through $(1, 7)$.

(A) Given the parametric equations $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$.
Differentiating with respect to $\theta$,we get $\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta$ and $\frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta$.
At $\theta = \frac{\pi}{4}$,the slope is $m = -\tan(\frac{\pi}{4}) = -1$.
The coordinates of the point at $\theta = \frac{\pi}{4}$ are $x = a \cos^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$ and $y = a \sin^3(\frac{\pi}{4}) = a(\frac{1}{\sqrt{2}})^3 = \frac{a}{2 \sqrt{2}}$.
The equation of the tangent line is $y - y_1 = m(x - x_1)$.
$y - \frac{a}{2 \sqrt{2}} = -1(x - \frac{a}{2 \sqrt{2}})$.
$y - \frac{a}{2 \sqrt{2}} = -x + \frac{a}{2 \sqrt{2}}$.
$x + y = \frac{a}{2 \sqrt{2}} + \frac{a}{2 \sqrt{2}} = \frac{2a}{2 \sqrt{2}} = \frac{a}{\sqrt{2}}$.

(C) Given the curve equation is $y^2 = px^3 + q \dots (i)$.
Since the point $(2, 3)$ lies on the curve,we substitute $x=2$ and $y=3$ into $(i)$:
$3^2 = p(2)^3 + q \Rightarrow 9 = 8p + q \dots (ii)$.
Differentiating both sides of $(i)$ with respect to $x$:
$2y \frac{dy}{dx} = 3px^2 \Rightarrow \frac{dy}{dx} = \frac{3px^2}{2y}$.
The slope of the tangent at $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$.
The given tangent line is $y = 4x - 5$,which has a slope of $4$.
Equating the slopes: $2p = 4 \Rightarrow p = 2$.
Substituting $p = 2$ into $(ii)$:
$9 = 8(2) + q \Rightarrow 9 = 16 + q \Rightarrow q = -7$.
Thus,the values are $p = 2$ and $q = -7$.

(C) Given the curve $y=ax^3+bx^2+cx+5$.
Since the curve touches the $X$-axis at $(-2,0)$,it passes through $(-2,0)$ and its derivative at $x=-2$ is $0$.
Substituting $(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \dots (i)$.
Also,the derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$.
At $x=-2$,$\frac{dy}{dx} = 0 \Rightarrow 3a(4) + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \dots (ii)$.
The curve cuts the $Y$-axis at $Q$. Putting $x=0$ in the curve equation,$y=5$,so $Q$ is $(0,5)$.
The gradient at $Q$ is $3$,so $\left(\frac{dy}{dx}\right)_{x=0} = 3 \Rightarrow c = 3$.
Substituting $c=3$ into $(i)$ and $(ii)$:
$8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \dots (iii)$.
$12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \dots (iv)$.
Subtracting $(iii)$ from $(iv)$: $4a = -2 \Rightarrow a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into $(iii)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$.
Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

(A) Given curves are $y=10-x^2$ and $y=2+x^2$.
To find the point of intersection,equate the two equations: $10-x^2 = 2+x^2 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
For $x=2$,$y=6$. For $x=-2$,$y=6$. Let us consider the point $(2, 6)$.
For the first curve $y=10-x^2$,the slope $m_1 = \frac{dy}{dx} = -2x$. At $x=2$,$m_1 = -4$.
For the second curve $y=2+x^2$,the slope $m_2 = \frac{dy}{dx} = 2x$. At $x=2$,$m_2 = 4$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-4 - 4}{1 + (-4)(4)} \right| = \left| \frac{-8}{1 - 16} \right| = \left| \frac{-8}{-15} \right| = \frac{8}{15}$.
Thus,$|\tan \theta| = \frac{8}{15}$.

(B) Given curve is $y=x \log x$ ... $(i)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 + \log x$.
The slope of the normal is given by $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{1+\log x}$.
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 the given line,their slopes must be equal:
$-\frac{1}{1+\log x} = 1$
$\Rightarrow 1+\log x = -1$
$\Rightarrow \log x = -2$
$\Rightarrow x = e^{-2}$.
Substituting $x = e^{-2}$ in $(i)$,we get $y = e^{-2} \log(e^{-2}) = e^{-2}(-2) = -2e^{-2}$.
The point of contact is $(e^{-2}, -2e^{-2})$.
The equation of the normal is $y - y_1 = m(x - x_1)$:
$y - (-2e^{-2}) = 1(x - e^{-2})$
$y + 2e^{-2} = x - e^{-2}$
$x - y = 3e^{-2}$.

(D) Given curve is $y=\sqrt{9-2x^2}$.
If ordinate and abscissa are equal,then $y=x$.
Substituting $y=x$ in the curve equation: $x^2 = 9 - 2x^2 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \pm \sqrt{3}$.
Since $y = \sqrt{9-2x^2}$,$y$ must be positive. Thus,$x$ must be $\sqrt{3}$ because if $x = -\sqrt{3}$,then $y = \sqrt{3}$,but $x \neq y$.
So,the point of contact is $(\sqrt{3}, \sqrt{3})$.
Differentiating $y^2 = 9 - 2x^2$ with respect to $x$: $2y \frac{dy}{dx} = -4x \Rightarrow \frac{dy}{dx} = -\frac{2x}{y}$.
At $(\sqrt{3}, \sqrt{3})$,the slope $m = -\frac{2(\sqrt{3})}{\sqrt{3}} = -2$.
The equation of the tangent is $y - \sqrt{3} = -2(x - \sqrt{3})$.
$y - \sqrt{3} = -2x + 2\sqrt{3} \Rightarrow 2x + y - 3\sqrt{3} = 0$.

(A) Given curves are $y=2x^2$ and $x=2y^2$.
For the curve $y=2x^2$,the slope of the tangent is $\frac{dy}{dx} = 4x$.
At the point $(1,1)$,the slope $m_1 = 4(1) = 4$.
For the curve $x=2y^2$,differentiating with respect to $x$,we get $1 = 4y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{4y}$.
At the point $(1,1)$,the slope $m_2 = \frac{1}{4(1)} = \frac{1}{4}$.
Let $\theta$ be the angle between the two tangents. The formula for the angle is $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{4 - 1/4}{1 + 4(1/4)} \right| = \left| \frac{15/4}{1 + 1} \right| = \frac{15/4}{2} = \frac{15}{8}$.
Therefore,$\theta = \tan^{-1}\left(\frac{15}{8}\right)$.

(B) Given the curve equation: $xy + ax + by = 0$.
Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$:
$1(1) + a(1) + b(1) = 0 \implies 1 + a + b = 0 \implies a + b = -1$ ... $(i)$
Now,differentiate the equation with respect to $x$:
$y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$
$\frac{dy}{dx}(x + b) = -(y + a)$
$\frac{dy}{dx} = -\frac{y + a}{x + b}$
Given the slope at $(1, 1)$ is $2$:
$2 = -\frac{1 + a}{1 + b}$
$2(1 + b) = -(1 + a)$
$2 + 2b = -1 - a $
$a + 2b = -3 ... (ii)$
Subtracting $(i)$ from $(ii)$:
$(a + 2b) - (a + b) = -3 - (-1)$
$b = -2$
Substituting $b = -2$ into $(i)$:
$a - 2 = -1 \implies a = 1$
Finally, calculate $3a + b$:
$3(1) + (-2) = 3 - 2 = 1$.

(B) The slope of the chord $AB$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-3)}{4 - 1} = \frac{6}{3} = 2$.
Since the tangent is parallel to the chord $AB$,the slope of the tangent must also be $2$.
Given the curve $y = x - \frac{4}{x}$,we find the derivative $\frac{dy}{dx} = 1 + \frac{4}{x^2}$.
Setting the derivative equal to the slope of the chord: $1 + \frac{4}{x^2} = 2$.
This simplifies to $\frac{4}{x^2} = 1$,which implies $x^2 = 4$,so $x = \pm 2$.
For $x = 2$,$y = 2 - \frac{4}{2} = 0$.
For $x = -2$,$y = -2 - \frac{4}{-2} = -2 + 2 = 0$.
Thus,the required points are $(2, 0)$ and $(-2, 0)$.

(C) Given the parametric equations of the curve are $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = 2(-\sin t + \sin t + t \cos t) = 2t \cos t$
$\frac{dy}{dt} = 2(\cos t - \cos t + t \sin t) = 2t \sin t$
Now,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t \sin t}{2t \cos t} = \tan t$.
The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{\tan t} = -\frac{\cos t}{\sin t}$.
The equation of the normal at point '$t$' is given by $y - y_1 = m_n(x - x_1)$:
$y - 2(\sin t - t \cos t) = -\frac{\cos t}{\sin t} [x - 2(\cos t + t \sin t)]$
Multiplying by $\sin t$:
$y \sin t - 2 \sin^2 t + 2t \sin t \cos t = -x \cos t + 2 \cos^2 t + 2t \sin t \cos t$
$x \cos t + y \sin t = 2(\sin^2 t + \cos^2 t)$
$x \cos t + y \sin t = 2$
The distance of the line $Ax + By + C = 0$ from the origin $(0,0)$ is given by $\frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$A = \cos t$,$B = \sin t$,and $C = -2$.
Distance $= \frac{|-2|}{\sqrt{\cos^2 t + \sin^2 t}} = \frac{2}{1} = 2$ units.

(C) $y^2=px^3+q$ ... $(i)$
Differentiating both sides with respect to $x$,we get
$2y \cdot \frac{dy}{dx} = 3px^2$
$\frac{dy}{dx} = \frac{3px^2}{2y}$
At the point $(2,3)$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{3p(2)^2}{2(3)} = \frac{12p}{6} = 2p$
The slope of the line $y=4x-5$ is $4$.
Since the line is tangent to the curve,their slopes must be equal:
$2p = 4 \Rightarrow p = 2$
Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we substitute the values:
$3^2 = p(2)^3 + q$
$9 = 8p + q$
Substituting $p=2$:
$9 = 8(2) + q$
$9 = 16 + q \Rightarrow q = -7$
Therefore,$p-q = 2 - (-7) = 2 + 7 = 9$.

(C) Given the curve $y = \frac{x}{x^2-3}$.
Finding the derivative: $\frac{dy}{dx} = \frac{(x^2-3)(1) - x(2x)}{(x^2-3)^2} = \frac{x^2-3-2x^2}{(x^2-3)^2} = \frac{-(x^2+3)}{(x^2-3)^2}$.
The slope of the line $2x + 6y - 11 = 0$ is $m = -\frac{2}{6} = -\frac{1}{3}$.
Since the tangent is parallel to the line,the slope of the tangent at $(\alpha, \beta)$ is $-\frac{1}{3}$.
So,$\frac{-(\alpha^2+3)}{(\alpha^2-3)^2} = -\frac{1}{3} \Rightarrow 3(\alpha^2+3) = (\alpha^2-3)^2$.
Let $t = \alpha^2$. Then $3t + 9 = t^2 - 6t + 9 \Rightarrow t^2 - 9t = 0$.
Since $(\alpha, \beta) \neq (0,0)$,$\alpha^2 \neq 0$,so $\alpha^2 = 9$,which means $\alpha = \pm 3$.
If $\alpha = 3$,$\beta = \frac{3}{3^2-3} = \frac{3}{6} = \frac{1}{2}$.
If $\alpha = -3$,$\beta = \frac{-3}{(-3)^2-3} = \frac{-3}{6} = -\frac{1}{2}$.
Now,calculate $|6\alpha + 2\beta|$:
For $(\alpha, \beta) = (3, 1/2)$,$|6(3) + 2(1/2)| = |18 + 1| = 19$.
For $(\alpha, \beta) = (-3, -1/2)$,$|6(-3) + 2(-1/2)| = |-18 - 1| = |-19| = 19$.
Thus,$|6\alpha + 2\beta| = 19$.

(B) The slope of the normal to the curve $y=f(x)$ at a point is given by $m_n = \tan(\theta)$,where $\theta$ is the angle made with the positive $X$-axis.
Given $\theta = \frac{3 \pi}{4}$,the slope of the normal is $m_n = \tan\left(\frac{3 \pi}{4}\right) = -1$.
We know that the slope of the normal is also related to the derivative of the function by the formula $m_n = -\frac{1}{f^{\prime}(x)}$.
At the point $(3,4)$,we have $m_n = -\frac{1}{f^{\prime}(3)}$.
Equating the two expressions for the slope of the normal:
$-1 = -\frac{1}{f^{\prime}(3)}$
$f^{\prime}(3) = 1$.

(A) Since the point $(2,3)$ lies on the curve $y^2=px^3+q$,we have:
$3^2 = p(2)^3 + q$
$9 = 8p + q$ ...$(i)$
Now,differentiate the curve equation $y^2=px^3+q$ with respect to $x$:
$2y \frac{dy}{dx} = 3px^2$
$\frac{dy}{dx} = \frac{3px^2}{2y}$
The slope of the tangent $y=4x-5$ is $4$. Therefore,the derivative at $(2,3)$ must be equal to $4$:
$\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3p(2)^2}{2(3)} = 4$
$\frac{12p}{6} = 4$
$2p = 4 \Rightarrow p = 2$ ...$(ii)$
Substitute $p=2$ into equation $(i)$:
$8(2) + q = 9$
$16 + q = 9$
$q = 9 - 16 = -7$
Thus,$p=2$ and $q=-7$.

(A) Given parametric equations are $x=t^2$ and $y=2t$.
First,find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$.
At $t=2$,the slope of the tangent is $m_T = \frac{1}{2}$.
The slope of the normal is $m_N = -\frac{1}{m_T} = -2$.
At $t=2$,the coordinates of the point are $x = (2)^2 = 4$ and $y = 2(2) = 4$.
The equation of the normal at $(4, 4)$ with slope $m_N = -2$ is given by $(y - y_1) = m_N(x - x_1)$.
Substituting the values: $(y - 4) = -2(x - 4)$.
$y - 4 = -2x + 8$.
$2x + y - 12 = 0$.

(A) Given the curve $y=x^3+ax-b$.
The slope of the tangent is $\frac{dy}{dx} = 3x^2+a$.
The slope of the line $y-x+4=0$ is $m_1 = 1$.
Since the tangent is perpendicular to the line,the slope of the tangent $m_2$ must satisfy $m_1 \times m_2 = -1$,so $m_2 = -1$.
At the point $(1,-5)$,$\frac{dy}{dx} = 3(1)^2+a = 3+a$.
Setting $3+a = -1$,we get $a = -4$.
Since the point $(1,-5)$ lies on the curve,we substitute $x=1, y=-5, a=-4$ into the equation:
$-5 = (1)^3 + (-4)(1) - b
-5 = 1 - 4 - b
-5 = -3 - b
b = 2$.
The equation of the curve is $y = x^3 - 4x - 2$.
Checking the options:
For $(2,-2)$: $y = (2)^3 - 4(2) - 2 = 8 - 8 - 2 = -2$.
Since the point $(2,-2)$ satisfies the equation,it lies on the curve.

(D) Given the curve $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$.
To find the slope of the tangent at $(a, b)$,we differentiate 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 have:
$n(1)^{n-1} \cdot \frac{1}{a} + n(1)^{n-1} \cdot \frac{1}{b} \cdot \frac{dy}{dx} = 0$.
$\frac{n}{a} + \frac{n}{b} \cdot \frac{dy}{dx} = 0$.
$\frac{dy}{dx} = -\frac{b}{a}$.
The equation of the line passing through $(a, b)$ with slope $m = -\frac{b}{a}$ is:
$y - b = -\frac{b}{a}(x - a)$.
$ay - ab = -bx + ab$.
$bx + ay = 2ab$.
Dividing by $ab$,we get:
$\frac{x}{a} + \frac{y}{b} = 2$.

(D) Given the curve equation is $y = 4xe^{x}$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$ using the product rule:
$\frac{dy}{dx} = 4(e^{x} + xe^{x}) = 4e^{x}(1 + x)$.
Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:
$\left(\frac{dy}{dx}\right)_{x=-1} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.
Since the slope of the tangent is $0$,the tangent is a horizontal line parallel to the $X$-axis.
The equation of a horizontal line passing through $\left(-1, -\frac{4}{e}\right)$ is $y = y_{1}$,which is $y = -\frac{4}{e}$.

(A) Given curve: $y = \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right)$.
First,find the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = \sqrt{2} \cos \left(2x + \frac{\pi}{4}\right) \cdot 2 = 2\sqrt{2} \cos \left(2x + \frac{\pi}{4}\right)$.
At $x = \frac{\pi}{4}$,the slope $m$ is:
$m = \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = 2\sqrt{2} \cos \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = 2\sqrt{2} \cos \left(\frac{3\pi}{4}\right) = 2\sqrt{2} \left(-\frac{1}{\sqrt{2}}\right) = -2$.
Now,find the $y$-coordinate at $x = \frac{\pi}{4}$:
$y = \sqrt{2} \sin \left(2 \cdot \frac{\pi}{4} + \frac{\pi}{4}\right) = \sqrt{2} \sin \left(\frac{3\pi}{4}\right) = \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = 1$.
The point of tangency is $\left(\frac{\pi}{4}, 1\right)$.
The equation of the tangent is $(y - y_1) = m(x - x_1)$:
$(y - 1) = -2 \left(x - \frac{\pi}{4}\right)$
$y - 1 = -2x + \frac{\pi}{2}$
$2x + y - \frac{\pi}{2} - 1 = 0$.

(D) Given the curve $y=ax^3+bx^2+cx+5$. Since it touches the $X$-axis at $P(-2,0)$,the point $(-2,0)$ lies on the curve and the gradient at this point is $0$.
Substituting $P(-2,0)$ into the equation: $0 = a(-8) + b(4) + c(-2) + 5 \Rightarrow -8a + 4b - 2c = -5 \Rightarrow 8a - 4b + 2c = 5 \quad (1)$.
The derivative is $\frac{dy}{dx} = 3ax^2 + 2bx + c$. At $P(-2,0)$,$\frac{dy}{dx} = 0$: $3a(-2)^2 + 2b(-2) + c = 0 \Rightarrow 12a - 4b + c = 0 \quad (2)$.
The curve cuts the $Y$-axis at $Q(0,k)$. At $Q$,$x=0$ and the gradient is $3$: $\frac{dy}{dx}|_{x=0} = 3(a)(0)^2 + 2(b)(0) + c = 3 \Rightarrow c = 3$.
Substituting $c=3$ into $(1)$ and $(2)$:
$(1): 8a - 4b + 6 = 5 \Rightarrow 8a - 4b = -1 \quad (3)$.
$(2): 12a - 4b + 3 = 0 \Rightarrow 12a - 4b = -3 \quad (4)$.
Subtracting $(3)$ from $(4)$: $(12a - 8a) = -3 - (-1) \Rightarrow 4a = -2 \Rightarrow a = -\frac{1}{2}$.
Substituting $a = -\frac{1}{2}$ into $(3)$: $8(-\frac{1}{2}) - 4b = -1 \Rightarrow -4 - 4b = -1 \Rightarrow -4b = 3 \Rightarrow b = -\frac{3}{4}$.
Thus,$a = -\frac{1}{2}, b = -\frac{3}{4}, c = 3$.

(D) Given the parametric equations of the curve: $x=4 \sec \theta$ and $y=4 \tan^2 \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = 4 \sec \theta \tan \theta$ and $\frac{dy}{d\theta} = 8 \tan \theta \sec^2 \theta$.
The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{8 \tan \theta \sec^2 \theta}{4 \sec \theta \tan \theta} = 2 \sec \theta$.
At $\theta = \frac{\pi}{4}$,the slope of the tangent is $2 \sec(\frac{\pi}{4}) = 2 \sqrt{2}$.
The slope of the normal is $-\frac{1}{\text{slope of tangent}} = -\frac{1}{2 \sqrt{2}}$.
At $\theta = \frac{\pi}{4}$,the point $(x, y)$ is $(4 \sec \frac{\pi}{4}, 4 \tan^2 \frac{\pi}{4}) = (4 \sqrt{2}, 4)$.
The equation of the normal is $y - y_1 = m_n(x - x_1)$:
$y - 4 = -\frac{1}{2 \sqrt{2}}(x - 4 \sqrt{2})$.
Multiplying by $2 \sqrt{2}$:
$2 \sqrt{2} y - 8 \sqrt{2} = -x + 4 \sqrt{2}$.
Rearranging gives $x + 2 \sqrt{2} y = 12 \sqrt{2}$.

(C) The slope of the line $6x - y - 4 = 0$ is $6$. Since this line is tangent to the curve $y^{2} = ax^{3} + b$ at the point $(1, 2)$,the derivative at this point must equal the slope of the line.
Differentiating the curve equation with respect to $x$:
$2y \frac{dy}{dx} = 3ax^{2} \Rightarrow \frac{dy}{dx} = \frac{3ax^{2}}{2y}$.
At the point $(1, 2)$,the slope is $\left(\frac{dy}{dx}\right)_{(1, 2)} = \frac{3a(1)^{2}}{2(2)} = \frac{3a}{4}$.
Equating the slope to $6$:
$\frac{3a}{4} = 6 \Rightarrow 3a = 24 \Rightarrow a = 8$.
Since the point $(1, 2)$ lies on the curve,it must satisfy the equation:
$(2)^{2} = a(1)^{3} + b \Rightarrow 4 = 8(1) + b \Rightarrow b = 4 - 8 = -4$.
Therefore,$a + b = 8 + (-4) = 4$.

(D) Given the curve equation: $2x^{2} + 3y^{2} - 5 = 0$.
Differentiating both sides with respect to $x$:
$4x + 6y \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y}$.
At the point $P(1, 1)$,the slope of the tangent $(m_{t})$ is:
$m_{t} = -\frac{2(1)}{3(1)} = -\frac{2}{3}$.
The slope of the normal $(m_{n})$ is the negative reciprocal of the tangent slope:
$m_{n} = -\frac{1}{m_{t}} = -\frac{1}{-2/3} = \frac{3}{2}$.
The equation of the normal at $(x_{1}, y_{1}) = (1, 1)$ is given by:
$y - y_{1} = m_{n}(x - x_{1})$
$y - 1 = \frac{3}{2}(x - 1)$
$2(y - 1) = 3(x - 1)$
$2y - 2 = 3x - 3$
$3x - 2y - 1 = 0$.

(C) Given the curve $y = \sin \left(\frac{\pi x}{4}\right)$.
First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:
$\frac{dy}{dx} = \frac{\pi}{4} \cos \left(\frac{\pi x}{4}\right)$.
At the point $(2, 1)$,the slope of the tangent $m_t$ is:
$m_t = \left(\frac{dy}{dx}\right)_{(2, 1)} = \frac{\pi}{4} \cos \left(\frac{2\pi}{4}\right) = \frac{\pi}{4} \cos \left(\frac{\pi}{2}\right) = \frac{\pi}{4} \times 0 = 0$.
Since the slope of the tangent is $0$,the tangent is a horizontal line (parallel to the $X$-axis).
Therefore,the normal,which is perpendicular to the tangent,must be a vertical line.
The equation of a vertical line passing through $(2, 1)$ is $x = 2$.
Thus,the correct option is $C$.

(D) Given the equation of the curve is $2x^{2} + y^{2} = 12$.
Differentiating with respect to $x$,we get $4x + 2y \frac{dy}{dx} = 0$.
Thus,$\frac{dy}{dx} = -\frac{2x}{y}$.
The slope of the tangent at $(2, 2)$ is $m_{t} = -\frac{2(2)}{2} = -2$.
The slope of the normal at $(2, 2)$ is $m_{n} = -\frac{1}{m_{t}} = -\frac{1}{-2} = \frac{1}{2}$.
The equation of the normal at $(2, 2)$ is given by $y - y_{1} = m_{n}(x - x_{1})$.
Substituting the values,$y - 2 = \frac{1}{2}(x - 2)$.
Multiplying by $2$,we get $2y - 4 = x - 2$.
Rearranging the terms,$x - 2y + 2 = 0$.
Therefore,the correct option is $D$.

(C) The slope of the line $y=4x-5$ is $4$.
Since the line touches the curve at $(2,3)$,the derivative of the curve at this point must equal the slope of the line.
Given $y^{2}=ax^{3}+b$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 3ax^{2}$.
Thus,$\frac{dy}{dx} = \frac{3ax^{2}}{2y}$.
At the point $(2,3)$,the slope is $\frac{3a(2)^{2}}{2(3)} = \frac{12a}{6} = 2a$.
Equating this to the slope of the line,we get $2a = 4$,which implies $a = 2$.
Since the point $(2,3)$ lies on the curve $y^{2}=ax^{3}+b$,we substitute $x=2, y=3$ and $a=2$ into the equation:
$3^{2} = 2(2)^{3} + b$
$9 = 2(8) + b$
$9 = 16 + b$
$b = 9 - 16 = -7$.
Therefore,$a=2$ and $b=-7$.

(A) Given the curve $y = \log_e x$.
First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at point $P(1, 0)$.
$\frac{dy}{dx} = \frac{1}{x}$.
At point $P(1, 0)$,the slope of the tangent $m_t = \left. \frac{dy}{dx} \right|_{(1, 0)} = \frac{1}{1} = 1$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\frac{1}{1} = -1$.
The equation of the normal line passing through $(x_1, y_1) = (1, 0)$ with slope $m_n = -1$ is:
$y - y_1 = m_n(x - x_1)$
$y - 0 = -1(x - 1)$
$y = -x + 1$
$x + y = 1$.

(A) Given curve: $y^2 = ax^3 + b$.
Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which simplifies to $9 = 8a + b$ (Equation $1$).
Differentiating the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$,so $\frac{dy}{dx} = \frac{3ax^2}{2y}$.
At the point $(2,3)$,the slope of the tangent is $\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$.
The given line is $y = 4x - 5$,which has a slope of $4$.
Equating the slopes: $2a = 4$,which gives $a = 2$.
Substituting $a = 2$ into Equation $1$: $9 = 8(2) + b \Rightarrow 9 = 16 + b \Rightarrow b = -7$.
Now,calculate $7a + 2b$: $7(2) + 2(-7) = 14 - 14 = 0$.

(C) Given the curve $y = \sqrt{x - 1}$.
Finding the derivative,we get $\frac{dy}{dx} = \frac{1}{2\sqrt{x - 1}}$. Let this be the slope of the tangent $m_1 = \frac{1}{2\sqrt{x - 1}}$.
The equation of the line is $2x + y - 5 = 0$,which can be written as $y = -2x + 5$. The slope of this line is $m_2 = -2$.
Since the tangent is perpendicular to the line,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
Substituting the values,we have $\left(\frac{1}{2\sqrt{x - 1}}\right) \times (-2) = -1$.
This simplifies to $\frac{-1}{\sqrt{x - 1}} = -1$,which implies $\sqrt{x - 1} = 1$.
Squaring both sides,we get $x - 1 = 1$,so $x = 2$.
Substituting $x = 2$ into the curve equation,$y = \sqrt{2 - 1} = \sqrt{1} = 1$.
Thus,the required point is $(2, 1)$.

(C) Given the curve equation: $y=c \cosh \left(\frac{x}{c}\right)$ ... $(i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = c \cdot \frac{1}{c} \cdot \sinh \left(\frac{x}{c}\right) = \sinh \left(\frac{x}{c}\right)$
The formula for the length of the normal is given by:
$L = |y| \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}$
Substituting the values:
$L = y \sqrt{1 + \sinh^{2} \left(\frac{x}{c}\right)}$
Using the identity $\cosh^{2} \theta - \sinh^{2} \theta = 1$,we have $1 + \sinh^{2} \theta = \cosh^{2} \theta$:
$L = y \sqrt{\cosh^{2} \left(\frac{x}{c}\right)}$
$L = y \cosh \left(\frac{x}{c}\right)$
From equation $(i)$,we know $\cosh \left(\frac{x}{c}\right) = \frac{y}{c}$:
$L = y \cdot \left(\frac{y}{c}\right) = \frac{y^{2}}{c}$
Thus,the length of the normal is $\frac{y^{2}}{c}$.

(D) Given the curve equation: $y^{2} = ax^{2} + b$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 2ax$
$\frac{dy}{dx} = \frac{ax}{y}$.
The slope of the tangent at the point $(2, 3)$ is $\left(\frac{dy}{dx}\right)_{(2,3)} = \frac{a(2)}{3} = \frac{2a}{3}$.
The given equation of the tangent is $y = 4x - 5$,which is in the form $y = mx + c$,so the slope $m = 4$.
Equating the slopes: $\frac{2a}{3} = 4 \Rightarrow 2a = 12 \Rightarrow a = 6$.
Since the point $(2, 3)$ lies on the curve,it must satisfy the equation $y^{2} = ax^{2} + b$:
$(3)^{2} = 6(2)^{2} + b$
$9 = 6(4) + b$
$9 = 24 + b$
$b = 9 - 24 = -15$.
Thus,the values are $a = 6$ and $b = -15$.

(B) Given curve is $y=4 x e^{x}$.
First,find the derivative $\frac{dy}{dx}$ using the product rule:
$\frac{dy}{dx} = 4e^{x} + 4x e^{x} = 4e^{x}(1+x)$.
Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:
$\left(\frac{dy}{dx}\right)_{(-1, -4/e)} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.
Since the slope is $0$,the tangent is a horizontal line.
The equation of the tangent line is given by $y - y_1 = m(x - x_1)$:
$y - (-\frac{4}{e}) = 0(x - (-1))$.
$y + \frac{4}{e} = 0$.
Therefore,$y = -\frac{4}{e}$.

(D) Given the curve equation: $y=x^{3}-3x^{2}-9x+5$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 3x^{2}-6x-9$.
Since the tangent is parallel to the $x$-axis,its slope must be zero:
$\frac{dy}{dx} = 0$.
Substituting the derivative:
$3x^{2}-6x-9 = 0$.
Dividing by $3$:
$x^{2}-2x-3 = 0$.
Factoring the quadratic equation:
$(x-3)(x+1) = 0$.
Thus,the values of $x$ (abscissae) are $x=3$ and $x=-1$.

(C) Given curves are $r_1 = \sin \theta + \cos \theta$ and $r_2 = 2 \sin \theta$.
To find the intersection,set $r_1 = r_2$:
$\sin \theta + \cos \theta = 2 \sin \theta \Rightarrow \cos \theta = \sin \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4}$.
For $r_1 = \sin \theta + \cos \theta$,$\frac{dr_1}{d\theta} = \cos \theta - \sin \theta$. Let $\phi_1$ be the angle between the tangent and the radius vector,then $\tan \phi_1 = \frac{r_1}{dr_1/d\theta} = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}$.
At $\theta = \frac{\pi}{4}$,$\tan \phi_1 = \frac{1/\sqrt{2} + 1/\sqrt{2}}{1/\sqrt{2} - 1/\sqrt{2}} = \frac{\sqrt{2}}{0} = \infty \Rightarrow \phi_1 = \frac{\pi}{2}$.
For $r_2 = 2 \sin \theta$,$\frac{dr_2}{d\theta} = 2 \cos \theta$. Then $\tan \phi_2 = \frac{r_2}{dr_2/d\theta} = \frac{2 \sin \theta}{2 \cos \theta} = \tan \theta$.
At $\theta = \frac{\pi}{4}$,$\tan \phi_2 = \tan(\frac{\pi}{4}) = 1 \Rightarrow \phi_2 = \frac{\pi}{4}$.
The angle of intersection $\psi = |\phi_1 - \phi_2| = |\frac{\pi}{2} - \frac{\pi}{4}| = \frac{\pi}{4}$.

(C) The curves $y = \sin x$ and $y = \cos x$ intersect where $\sin x = \cos x$,which implies $\tan x = 1$. Since $0 < x < \frac{\pi}{2}$,the intersection occurs at $x = \frac{\pi}{4}$.
Let $m_1$ and $m_2$ be the slopes of the tangents to the curves at $x = \frac{\pi}{4}$.
For $y = \sin x$,$\frac{dy}{dx} = \cos x$. Thus,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
For $y = \cos x$,$\frac{dy}{dx} = -\sin x$. Thus,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values: $\tan \theta = |\frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})}| = |\frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}}| = |\frac{\sqrt{2}}{\frac{1}{2}}| = 2\sqrt{2}$.
Therefore,$\theta = \tan^{-1}(2\sqrt{2})$.