Question

If $M(x_0,y_0)$ is the point on the curve $3x^2-4y^2=72,$ which is nearest to the line $3x+2y+1=0,$ then the value of $(x_0+y_0)$ is equal to

• A. $3$
• B. $-3$
• C. $9$
• D. $-9$

Answer 1

The correct option is B $-3$

Slope of given line $=\frac{-3}{2}$
First of all, we try to locate the points on the curve at which the tangent is parallel to the given line.
So, differentiating both sides with respect to $x$ of $3x^2-4y^2=72,$ we get
$\frac{dy}{dx}=\frac{3x}{4y}$
$\therefore \frac{3x}{4y}=\frac{-3}{2}$
$\Rightarrow x=-2y$

Now, $3{(-2y)}^2-4y^2=72$
$\Rightarrow y^2=9$
$\Rightarrow y=-3,3$
So, points are $(-6,3)$ and $(6,-3).$

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 $M(-6,3)=(x_0,y_0)$
Hence, $(x_0+y_0)=-6+3=-3$