Question

The equation of those tangents to $4x^2-9y^2=36$ which are perpendicular to the straight line $5x+2y-10=0$ are

• A. $5(y-3)=2(x-\frac{\sqrt{117}}{2})$
• B. $2x-5y+10-2\sqrt{18}=0$
• C. $2x-5y-10-2\sqrt{18}=0$
• D. None of the above

Answer 1

The correct option is D None of the above

We have, $4x^2-9y^2=36$
On differentiating w.r.t $x$ we get
$8x-18y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}$
Slope of the tangent $\frac{4x}{9y}$
for this tangent to be perpendicular to the straight line $5x+2y-10=0$, we must have
$\frac{4x}{9y}\times (-\frac{5}{2})=-1$
$\Rightarrow y=\frac{10x}{9}$
Putting this value of $y$ in $4x^2-9y^2=36$, we get
$64x^2=324$, which does not have real roots. Hence, at no points on the given curve can the tangent be perpendicular to given line.