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  • Slope of the normal chord of that gets bisected at <img alt="\mathrm{(8,2)}

Slope of the normal chord of \mathrm{y^{2}=8 x} that gets bisected at \mathrm{(8,2)} is

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

Let \mathrm{A B} be the normal chord where

\mathrm{A \equiv\left(2^{t_{1}^{2}}, 4^{t_{1}}\right), B \equiv\left(2^{t_{2}^{2}}, 4^{t_{2}}\right),}

\mathrm{It's \; slope =\frac{2}{t_{1}+t_{2}}}
We also have
\mathrm{t_{2}=-t_{1}-\frac{2}{t_{1}}}

and 16=2\left(\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}\right), 4=4\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)
\Rightarrow \quad \mathrm{t}_{1}+\mathrm{t}_{2}=1
Thus slope is 2 .

Posted by

shivangi.shekhar

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