Question

The length of the chord of the parabola $y^2=4\text{ax}(a>0)$ 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

Answer: (B)

Equation of OP:-
$y=x\tan \alpha$
Solving with $y^2=4ax$, we get :
$x^2{\tan }^2\alpha =4ax\Rightarrow x=4a{\cot }^2\alpha$
Substituting, $y=4a\cot \alpha$
$\therefore P=\left(4a{\cot }^2\alpha ,4a\cot \alpha \right)$
So, $\overline{OP}=\sqrt{16a^2{\cot }^4\alpha +16a^2{\cot }^2\alpha }=4a\cot \alpha \text{cosec}\alpha$
$($ as $0^{\circ }<\alpha <90^{\circ }$, so, $\cot a>0,\text{cosec}\alpha >0)$