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

Option: 1

1


Option: 2

-1


Option: 3

2


Option: 4

-2


Answers (1)

best_answer

Let \mathrm{AB} be the normal chord where \mathrm{\mathrm{A} \equiv\left(2 \mathrm{t}_1{ }^2, 4 \mathrm{t}_1\right), \mathrm{B} \equiv\left(2 \mathrm{t}_2{ }^2, 4 \mathrm{t}_2\right)}. It's slope \mathrm{=\frac{2}{\mathrm{t}_1+\mathrm{t}_2}}  We also have \mathrm{\mathrm{t}_2=-\mathrm{t}_1-\frac{2}{\mathrm{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 \mathrm{t}_1+\mathrm{t}_2=1}. Thus slope is 2.

Hence option 2 is correct.

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