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

• A

  $\pm 4 a \cot \alpha \operatorname{cosec} \alpha$
• Correct answer
  B

  $4 a \cot \alpha \operatorname{cosec} \alpha$
• C

  $-4 a \cot \alpha \operatorname{cosec} \alpha$
• D

  $4 a \operatorname{cosec}^{2} \alpha$

Solution

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

Hence, the correct answer is option: 'B'.