Question

There are two perpendicular straight lines touching the parabola $y^2=4a(x+a)$ and $y^2=4b(x+b)$, then the point of intersection of these two lines lie on the line given by

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

Answer 1

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

the equation of the tangent of slope $m_1$ to $y^2=4a(x+a)$ is $y=m_1(x+a)+\frac{a}{m_1}$.........$\left(1\right)$
The equation of tangent of slope $m_2$ to the parabola $y^2=4b(x+b)$ is $y=m_2(x+b)+\frac{b}{m_2}$.........$\left(2\right)$
We know that, $m_1=-\frac{1}{m_2}$
$\Rightarrow y=-\frac{1}{m_1}(x+b)-b{m}_1$..........$\left(3\right)$
$\left(1\right)-\left(3\right)$ gives
$0=(m_1+\frac{1}{m_1})x+a(m_1+\frac{1}{m_1})+b(m_1+\frac{1}{m_1})$
$0=(x+a+b)(m_1+\frac{1}{m_1})$
$x+a+b=0$
This is the equation of line on which point of intersection lies.