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If a normal chord of \mathrm{y}^2=4 \mathrm{ax} subtends an angle \mathrm{ \pi / 2} at the vertex of the parabola then it's slope is equal to
 

Option: 1

\pm 1


 


Option: 2

\pm \sqrt{2}
 


Option: 3

\pm 2
 


Option: 4

\text{none of these}


Answers (1)

best_answer

Let \mathrm{AB} be a normal chord, where \mathrm{\mathrm{A} \equiv\left(\mathrm{at}_1{ }^2, 2 \mathrm{at}_1\right) and \mathrm{B} \equiv\left(\mathrm{at}_2{ }^2, 2 \mathrm{at} \mathrm{t}_2\right)}

We have \mathrm{\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}\: and \: \mathrm{t}_1 \mathrm{t}_2=4 \Rightarrow \mathrm{t}_1 \mathrm{t}_2=-\mathrm{t}_1{ }^2-2=-4 \Rightarrow \mathrm{t}_1{ }^2=2.}

Now slope of chord \mathrm{\mathrm{AB}=\frac{2}{\mathrm{t}_1+\mathrm{t}_2}=-\mathrm{t}_1= \pm \sqrt{2}}

Hence option 2 is correct.

Posted by

Ritika Harsh

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