Question

The tangent and normal at the point $P(a{t}^2,2at)$ to the parabola $y^2=4ax$ meet the $x$-axis at $T$ and $G$ respectively. Then the angle at which the tangent at $P$ to the parabola is inclined to the tangent at $P$ to the circle through $T,P,G$ is

• A. ${\tan }^{-1}{t}^2$
• B. ${\cot }^{-1}{t}^2$
• C. ${\tan }^{-1}\pm t$
• D. ${\cot }^{-1}t$

Answer 1

Answer: (C)

${\tan }^{-1}\pm t$

Given, a parabola $y^2=4ax$ , and a point $P(a{t}^2,2at)$ be a point on the parabola.
Focus $S$ of parabola is $(a,0)$, which is also the midpoint of $TG$. (It can be shown to be always true for a parabola)
Hence, $S$ is the center of the rightangled triangle $PTG$
Slope of tangent at $P$ on the parabola is $m_1=\frac{1}{t}$
Now, slope of radius $PS=\frac{2at}{a{t}^2-a}=\frac{2t}{t^2-1}$
Slope of tangent at $P$ on the circle is $m_2=\frac{1-t^2}{2t}(\because$ radius and tangent are perpendicular to each other.)
Now, let $\theta$ be the angle between the tangent at $P$ to the parabola and tangent at $P$ to the circle.
$\tan \theta =\left|\frac{\frac{1}{t}-\frac{1-t^2}{2t}}{\frac{1}{t}\times \frac{1-t^2}{2t}}\right|$
$\tan \theta =\pm t$
$\Rightarrow \theta ={\tan }^{-1}(\pm t)$