Question

Let $L_1$ be a tangent to the parabola$y^{{}_2}=4(x+1)$ and $L_2$ be a tangent to the parabola $y^2=8(x+2)$ such that $L_1\text{and}{L}_2$ intersect at right angles. Then $L_1\text{and}{L}_2$ meet on the straight line:

• A. $x+2y=0$
• B. $x+2=0$
• C. $2x+1=0$
• D. $x+3=0$

Answer 1

The correct option is D

$x+3=0$

Explanation for correct option:

Finding the equation for a straight line:

Let $t_1\text{be the}\text{tangent of}{y}^2=4(x+1)$

$t_{1}y=(x+1)+{t_1}^2\dots \dots .\left(i\right)$

Similarly,

$$\begin{gathered} t_2\text{tangent of}{y}^2=8(x+2) \\ {t}_{2}y=(x+2)+2{t_2}^2\dots .\left(ii\right) \end{gathered}$$

Given $L_1\perp L_2$

$$\begin{gathered} \left(\frac{1}{t_1}\right)\left(\frac{1}{t_2}\right)=-1 \\ \Rightarrow t_1{t}_2=-1 \end{gathered}$$

Multiplying equation $\left(i\right)\text{by}{t}_2$ and $\left(ii\right)\text{by}{t}_1$

$$\begin{gathered} t_1{t}_{2}y=t_2(x+1)+t_2.{t_1}^2 \\ {t}_1{t}_{2}y=t_1(x+2)+2{t_2}^2.t_1 \\ \mathrm{Subtracting}\mathrm{above}\mathrm{equations},\mathrm{we}\mathrm{get} \\ (t_2-t_1)x+(t_2-2t_1)+t_2{t}_1(t_1-2t_2)=0 \\ (t_2–t_1)x+(t_2-2t_1)-(t_1-2t_2)=0 \\ (t_2-t_1)x+3t_2-3t_1=0 \\ (t_2-t_1)x+3(t_2-t_1)=0 \\ (t_2-t_1)\left(x+3\right)=0 \\ \Rightarrow x+3=0 \end{gathered}$$

Hence Option (D) is the correct answer.