Practicing Success
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 |
$5(y-3)=2\left(x-\frac{\sqrt{117}}{2}\right)$ $2 x-5 y+10-2 \sqrt{18}=0$ $2 x-5 y-10-2 \sqrt{18}=0$ none of these |
none of these |
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. |