Question

A tangent and normal is drawn at the point $P=(16,16)$ of the parabola $y^2=16x$ which cut the axis of the parabola at the points $A$ and $B$, respectively. If the centre of the circle through $P$. $A$ and $B$ is $C$, then the angle between $PC$ and the axis of $x$ is

• A. ${\tan }^{-1}\frac{1}{2}$
• B. ${\tan }^{-1}2$
• C. ${\tan }^{-1}\frac{3}{4}$
• D. ${\tan }^{-1}\frac{4}{3}$

Answer 1

The correct option is D ${\tan }^{-1}\frac{4}{3}$

Given the parabola is $y^2=16x$.........(1).

To find the equation of the tangent and the normal at $P(16,16)$ differentiate (1) w.r.to $x$ we get,

$2y\frac{dy}{dx}$ $=16$

or, $\frac{dy}{dx}=\frac{2}{\sqrt{x}}$

or, $\frac{dy}{dx}{|}_{(16,16)}=\frac{2}{\sqrt{16}}=\frac{1}{2}$.

$\therefore$ the equation of tangent at $(16,16)$ to the parabola is

$(y-16)=\frac{1}{2}$$(x-16)$

or, $2y-32=x-16$

or, $x-2y=-16$

or, $\frac{x}{-16}+\frac{y}{8}$ $=1$,

so, co-ordinate of $A$ is $(-16,0)$.

Now, the equation of normal at $(16,16)$ to the parabola is

$(y-16)=(-2)(x-16)$

or, $2x+y=48$

or, $\frac{x}{24}+\frac{y}{48}$ $=1$,

so , co-ordinate of $B$ is $(24,0)$.

For, the circle which passes through $P$, $A$ and $B$, we claim that $AB$ is the diametre of it. As $AB$ subtend a right-angle at $P$. [Since angle between the tangent and the normal is ${90}^{\circ }$.]

$\therefore$ centre of the circle is $C=\left(\frac{24-16}{2},0\right)=(4,0)$.

$\therefore$ the angle made by the $PC$ with $x$-axis is

${\tan }^{-1}\frac{16-0}{16-4}$ $=$${\tan }^{-1}\frac{4}{3}$