The equations of those tangents to $4 x^2-9 y^2=36$ which are perpendicular to the straight line $5 x+2 y-10=0$, are
Answer & explanation
Correct answer: option 4
We have,
$4 x^2-9 y^2=36 \Rightarrow 8 x-18 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{4 x}{9 y}$
∴ Slope of the tangent $=\frac{4 x}{9 y}$
For this tangent to be perpendicular to the straight line $5 x+2 y-10=0$, we must have
$\frac{4 x}{9 y} \times\left(-\frac{5}{2}\right)=-1 \Rightarrow y=\frac{10 x}{9}$
Putting this value of y in $4 x^2-9 y^2=36$, we get $-64 x^2=324$, which does not have real roots. Hence, at no point on the given curve can the tangent be perpendicular to the given line.