Question

Two straight lines are perpendicular to each other one of them touches the parabola $y^2=4a(x+a)$ and the other touches $y^2=4b(x+b).$ The locus of the point of intersection of these two lines is

• A. $x+a+b=0$
• B. $x-a-b=0$
• C. $x=0$
• D. $x+a=0$

Answer 1

The correct option is A $x+a+b=0$

Let $m$ be the slope of one line, then slope of other line will be $-\frac{1}{m}$
Equation of the line touching the parabola $y^2=4a(x+a)$ is
$y=m(x+a)+\frac{a}{m}\cdots \left(1\right)$
Equation of the line touching the parabola $y^2=4b(x+b)$ is
$y=-\frac{1}{m}(x+b)+\frac{b}{-\frac{1}{m}}$
$y=-\frac{1}{m}(x+b)-bm\cdots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right)$ we get
$\frac{a}{m}+m(x+a)=-bm-\frac{1}{m}(x+b)$
$\Rightarrow x(m+\frac{1}{m})+a(m+\frac{1}{m})+b(m+\frac{1}{m})=0$
$\therefore x+a+b=0[\because m+\frac{1}{m}\ne 0]$