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Question: The point of the curve 3x<sup>2</sup> –4y<sup>2</sup> = 72 which is nearest to the line 3x + 2y – 1 ...

The point of the curve 3x2 –4y2 = 72 which is nearest to the line 3x + 2y – 1 = 0 is:

A

(+6, 3)

B

(6, –3)

C

(6, 6)

D

(6, 5)

Answer

(6, –3)

Explanation

Solution

For the nearest point on the curve, tangent drawn to curve at that point should be parallel to the given line

\ 6x1 – 8y1 (dydx)(x1,y1)\left( \frac{dy}{dx} \right)_{(x_{1},y_{1})}= 0

Ž (dydx)\left( \frac{dy}{dx} \right)= +34\frac{3}{4}. x1y1\frac{x_{1}}{y_{1}}= –32\frac{3}{2}

Ž x1 = –2y1

which satisfy 3x2 –4y2 = 72

Ž 12y12 –4y12 = 72

Ž y12 = 9

Ž y1 = ± 3

Ž x1 = ± 6

Hence required points are (–6, 3) and (6, –3).