If Mxo,yo is the point on the curve 3x2−4y2=72 which is nearest to the line 3x+2y+1=0, then the value of xo+yo is equal to

Slope of the give line =−32 The points on the curve at which the tangent is parallel to the given line. So,  differentiating both sides with respect to x of 3x2−4y2=72 we get dydx=3x4y=−32( given )⇒xy=−2 Now 3xy2−4=72y2⇒y2=9⇒y=−3,3 So, points are (−6,3) and (6,−3) Now, distance of (−6,3) from the given line =−18+6+113=1113 And distance of (6,−3) from the given line =18−6+113=1313 Clearly, the required point is on (−6,3)=x0,yo (given)  So, xo=−6,yo=3 Hence x0+y0=−6+3=−3