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Double ordinates \mathrm{AB} of the parabola \mathrm{y}^2=4 \mathrm{ax} subtends an angle \mathrm{\pi / 2} at the focus of the parabola then tangents drawn to parabola at \mathrm{\mathrm{A}\: and \: \mathrm{B}} will intersect at
 

Option: 1

(-4 \mathrm{a}, 0)

 


Option: 2

(-2 \mathrm{a}, 0)
 


Option: 3

(-3 \mathrm{a}, 0)
 


Option: 4

\text{None of these}


Answers (1)

best_answer

Let \mathrm{\mathrm{A} \equiv\left(\mathrm{at}^2, 2 \mathrm{at}\right), \mathrm{B} \equiv\left(\mathrm{at}^2,-2 \mathrm{at}\right) \cdot \mathrm{m}_{\mathrm{OA}}=\frac{2}{\mathrm{t}}, \mathrm{m}_{\mathrm{OB}}=-\frac{2}{\mathrm{t}}\: Thus\: \frac{2}{\mathrm{t}} \cdot-\frac{2}{\mathrm{t}}=-1 \Rightarrow \mathrm{t}^2=4}

Thus, tangents will intersect at \mathrm{(-4 a, 0).}

Hence option 1 is correct.

 

 

Posted by

sudhir.kumar

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