Find the point on the curve `3x^2-4y^2=72` which is nearest to the line `3x+2y+1=0.`

Options

A. (6, 3)
B. `(-6, -3)`
C. `(-6, 3)`
D. `(6, -3)`

Answer

C

Explanation

Point P is nearest to the given line if the tangent at P is parallel to the given line.
Now, the slope of tangent at `P(x_(1),y_(1))` is
`((dy)/(dx))_((x_(1)","y_(1)))=(18x_(1))/(24y_(1))=(3)/(4)(x_(1))/(y_(1))` which must be equal to `-3//2.`
Therefore,
`(3)/(4)(x_(1))/(y_(1))=-(3)/(2)`
`"or "x_(1)=-2y_(1)" (1)"`

Also, `(x_(1),y_(1))` lies on the curve. Hence,
`(x_(1)^(2))/(24)-(y_(1)^(2))/(18)=1" (2)"`
Solving (1) and (2), we get two points (6, -3) and (-6, 3) of which (-6, 3) is the nearest.