First slide

Tangents and normals

Question

$If M (x_{o}, y_{o}) is the point on the curve 3 x^{2} - 4 y^{2} = 72 which is nearest to the line$
$3 x + 2 y + 1 = 0, then the value of (x_{o} + y_{o}) is equal to$

Moderate

A

3
B

-3
C

9
D

-9

Solution

$Slope of the give line = - \frac{3}{2}$
$The points on the curve at which the tangent is parallel to the given line. So,$
$differentiating both sides with respect to x of 3 x^{2} - 4 y^{2} = 72 we get$
$\begin{array}{l} \frac{d y}{d x} = \frac{3 x}{4 y} = \frac{- 3}{2} (given) \\ \Rightarrow \frac{x}{y} = - 2 \\ Now 3 {(\frac{x}{y})}^{2} - 4 = \frac{72}{y^{2}} \Rightarrow y^{2} = 9 \Rightarrow y = - 3, 3 \\ So, points are (- 6, 3) and (6, - 3) \end{array}$
$Now, distance of (- 6, 3) from the given line = |\frac{- 18 + 6 + 1}{\sqrt{13}}| = \frac{11}{\sqrt{13}}$
$And distance of (6, - 3) from the given line = |\frac{18 - 6 + 1}{\sqrt{13}}| = \frac{13}{\sqrt{13}}$
$Clearly, the required point is on (- 6, 3) = (x_{0}, y_{o}) (given)$
$\begin{array}{l} So, x_{o} = - 6, y_{o} = 3 \\ Hence (x_{0} + y_{0}) = - 6 + 3 = - 3 \end{array}$

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