Q96 / 155 · WBJEE 2022 · Mathematics

Let the tangent and normal at any point $P\left(at^2\right.$, 2at), $(a > 0)$, on the parabola $y^2=4 a x$ meet the axis of the parabola at $T$ and $\mathrm{G}$ respectively. Then the radius of the circle through $\mathrm{P}, \mathrm{T}$ and $\mathrm{G}$ is

  • Correct answer
    A
    $a\left(1+t^2\right)$
  • B
    $\left(1+t^2\right)$
  • C
    $a\left(1-t^2\right)$
  • D
    $\left(1-\mathrm{t}^2\right)$

Solution

$P\left(a t^2, 2 a t\right), T\left(-a t^2, 0\right), G\left(2 a+a t^2, 0\right)$ Slope of $P T \times$ Slope of $P G=-1$ $\therefore$ TG is a diameter of the circle through points $\mathrm{P}, \mathrm{T}$ and $\mathrm{G}$. $\therefore \text { radius }=\frac{1}{2} \mathrm{TG}=\mathrm{a}\left(1+\mathrm{t}^2\right)$

Hence, the correct answer is option: 'A'.

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