Question

The length of the chord of the parabola $y^2=4ax(a>0)$ which passes through the vertex and makes an acute angle $\alpha$ with the axis of the parabola is

• A. $\pm 4a\cot \alpha \text{cosec}\alpha$
• B. $4a\cot \alpha \text{cosec}\alpha$
• C. $-4a\cot \alpha \text{cosec}\alpha$
• D. $4a{\text{cosec}}^2\alpha$

Answer 1

The correct option is B $4a\cot \alpha \text{cosec}\alpha$

$\angle AOX=\alpha$
Equation $AO$ is $y-0=\tan \alpha (x-0)$
$\Rightarrow y=x\tan \alpha$
$\because y^2=4ax$
$\Rightarrow x^2{\tan }^2\alpha =4ax$
$\Rightarrow x{\tan }^2\alpha =4a$
$\Rightarrow x=4a{\cot }^2\alpha$
$\therefore y=4a{\cot }^2\alpha \tan \alpha$
$\Rightarrow y=4a\cot \alpha$

$\therefore A=(4a{\cot }^2\alpha ,4a\cot \alpha )$
$\Rightarrow AO=\sqrt{16a^2{\cot }^4\alpha +16a^2{\cot }^2\alpha }$
$=4a\cot \alpha \sqrt{{\cot }^2\alpha +1}$
$=4a\cot \alpha \text{cosec}\alpha$