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An equilateral triangle $\mathrm{SAB}$ is inscribed in the parabola $\mathrm{\mathrm{y}^2=4 \mathrm{ax}}$ having it's focus at $\mathrm{' \mathrm{S}'}$ . If chord $\mathrm{\mathrm{AB}}$ lies towards the left of $\mathrm{S}$ , then side length of this triangle is

Option: 1
$3 \mathrm{a}(2-\sqrt{3})$

Option: 2
$4 \mathrm{a}(2-\sqrt{3})$

Option: 3
$2(2-\sqrt{3})$

Option: 4
$8 \mathrm{a}(2-\sqrt{3})$

Answers (1)

best_answer

Let $\mathrm{A} \equiv\left(\mathrm{at}_1^2, 2 \mathrm{at}_1\right), \mathrm{B} \equiv\left(\mathrm{at}_2^2,-2 \mathrm{at}_1\right).$ We have $\mathrm{ \mathrm{m}_{\mathrm{AS}}=\tan \left(\frac{\pi}{6}\right) \Rightarrow \frac{2 \mathrm{at}_1}{\mathrm{at}_1^2-\mathrm{a}}=-\frac{1}{\sqrt{3}}}$

$\mathrm{ \Rightarrow \quad \mathrm{t}_1{ }^2+2 \sqrt{3} \mathrm{t}_1=-1=0 \Rightarrow \mathrm{t}_1=-\sqrt{3} \pm 2 }$

Clearly $\mathrm{ \mathrm{t}_1=-\sqrt{3}-2 }$ is rejected. Thus $\mathrm{ \mathrm{t}_1=(2-\sqrt{3}) }$ . Hence $\mathrm{ \mathrm{AB}=4 \mathrm{at}_1=4 \mathrm{a}(2-\sqrt{3}) }$

Hence option 2 is correct.

Posted by

Gautam harsolia

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