Question

The equation of the line passing through the point $(x\text{'},y\text{'})$ and perpendicular to the line $yy\text{'}=2a(x+x\text{'})$ is

• A. $xy\text{'}+2ay+2ay\text{'}–x\text{'}y\text{'}=0$
• B. $xy\text{'}+2ay–2ay\text{'}–x\text{'}y\text{'}=0$
• C. $xy\text{'}+2ay+2ay\text{'}+x\text{'}y\text{'}=0$
• D. $xy\text{'}+2ay–2ay\text{'}+x\text{'}y\text{'}=0$

Answer 1

The correct option is B

$xy\text{'}+2ay–2ay\text{'}–x\text{'}y\text{'}=0$

Explanation of the correct option:

Given : Equation of the line is $yy\text{'}=2a(x+x\text{'})$………$\left(1\right)$

It can be written as, $y=\frac{2a}{y\text{'}}x+\frac{2a}{y\text{'}}x\text{'}$

Slope of the line is $\frac{2a}{y\text{'}}$.

Therefore, slope of the perpendicular line is $-\frac{y\text{'}}{2a}$.

We know that equation of the line is given as,

$y-y_1=m(x-x_1)$

$\Rightarrow$ $\left(y-y\text{'}\right)=\left(-\frac{y\text{'}}{2a}\right)\left(x-x\text{'}\right)$ $\left[\because x_1=x\text{'},y_1=y\text{'},m=-\frac{y\text{'}}{2a}\right]$

$\Rightarrow$ $2ay-2ay\text{'}=-xy\text{'}+x\text{'}y\text{'}$

$\Rightarrow$$xy\text{'}+2ay-2ay\text{'}-x\text{'}y\text{'}=0$

Hence, Option $\left(\mathrm{B}\right)$ is the correct answer.