Question

Locus of the point of intersection of perpendicular tangents drawn one each to the parabolas $y^2=4(x+1),y^2=8(x+2)$ is

• A. $x+12=0$
• B. $x+8=0$
• C. $x+4=0$
• D. $x+3=0$

Answer 1

Answer: (D)

$x+3=0$

Given parabolas
$y^2=4(x+1)$ ....(1)
$y^2=8(x+2)$ ....(2)
Let the point of intersection of tangents to the parabolas be $P(x_1,y_1)$
Differentiating (1) w.r.t x,
$2y\frac{dy}{dx}=4$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
Slope of tangent to (1) at $P(x_1,y_1)=\frac{2}{y_1}$
Differentiating (2) w.r.t x,
$2y\frac{dy}{dx}=8$
$\Rightarrow \frac{dy}{dx}=\frac{4}{y}$
Slope of tangent to (2) at $P(x_1,y_1)=\frac{4}{y_1}$
Since, the tangents are perpendicular
$\Rightarrow \frac{2}{y_1}\frac{4}{y_1}=-1$
$\Rightarrow y_1^2=-8$
$\Rightarrow 8(x+2)=-8$ (by 2)
$\Rightarrow x+3=0$
which is the required locus.