Q94 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
$S \equiv y^2-4 a x=0, S^{\prime} \equiv y^2+a x=0$ are two parabolas and $P(t)$ is a point on the parabola $\mathrm{S}^{\prime}=0$. If A and B are the feet of the perpendiculars from P on to coordinate axes and AB is a tangent to the parabola $\mathrm{S}=0$ at the point $\mathrm{Q}\left(t_1\right)$, then $t_1=$
- A$t$
- B$\frac{t}{4}$
- C$\frac{3 t}{4}$
- Correct answerD$\frac{t}{2}$
Solution
$\begin{aligned} & \text { } \mathrm{S} \equiv y^2-4 a x=0 \\ & \mathbf{S}^{\prime} \equiv y^2+4 a x=0\end{aligned}$
Let $\mathrm{P}(t)=\left(\frac{-t^2}{4 a}, t\right) ; \mathrm{So}, \mathrm{A}=\left(\frac{-t^2}{4 a}, 0\right)$ and $\mathrm{B}=(0, \mathrm{t})$ Line $\mathrm{AB}: 4 a x-t y+t=0$..(i) Consider, $\mathrm{Q}\left(t_1\right)=\left(\frac{t_1}{4 a}, \frac{t_1+t}{t}\right)$ Tangent at $\mathrm{Q}\left(t_1\right)$ of $S=0$ is $y\left(\frac{t_1+t}{t}\right)-2 a\left(x+\frac{t_1}{4 a}\right)=0$
$\Rightarrow 2 a x-y\left(1+\frac{t_1}{t}\right)+\frac{t_1}{2}=0$..(ii) On comparing (i) and (ii) we get $t_1=\frac{t}{2}$
Let $\mathrm{P}(t)=\left(\frac{-t^2}{4 a}, t\right) ; \mathrm{So}, \mathrm{A}=\left(\frac{-t^2}{4 a}, 0\right)$ and $\mathrm{B}=(0, \mathrm{t})$ Line $\mathrm{AB}: 4 a x-t y+t=0$..(i) Consider, $\mathrm{Q}\left(t_1\right)=\left(\frac{t_1}{4 a}, \frac{t_1+t}{t}\right)$ Tangent at $\mathrm{Q}\left(t_1\right)$ of $S=0$ is $y\left(\frac{t_1+t}{t}\right)-2 a\left(x+\frac{t_1}{4 a}\right)=0$
$\Rightarrow 2 a x-y\left(1+\frac{t_1}{t}\right)+\frac{t_1}{2}=0$..(ii) On comparing (i) and (ii) we get $t_1=\frac{t}{2}$
Hence, the correct answer is option: 'D'.