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The point on the parabola y^2= 64x which...

The point on the parabola `y^2= 64x` which is nearest to the line `4x +3y +35=0 `has coordinates

A

`(9,24)`

B

`(1, 81)`

C

`(4,-16)`

D

`(-9, -24)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given equation of parabola is
`y^(2)=64x " "…(i)`
The point at which the tangent to the curce is parallel to the line is the nearest point on the curve. On differentiating both sides of Eq. (i), we get
`2y(dy)/(dx)=64implies(dy)/(dx)=32/y`
Also, slope of the given line is `-4//3`
`therefore -4/3=32/y`
`impliesy=-24`
From Eq. (i), `(-24)^(2)=64x`
`impliesx=9`
Hence the required point is `(9,-24).`
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