Question

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

• A. $5\left(y-3\right)=4\left(x-\frac{\sqrt{11}}{2}\right)$
• B. $2x-5y+10-12\sqrt{3}=0$
• C. $2x-5y-10+12\sqrt{3}=0$
• D. None of these

Answer 1

Answer: (D)

$4x^2-9y^2=36\cdots \left(1\right)$
$\Rightarrow 8x-18y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}$
$\therefore$ Slope of the tangent $=\frac{4x}{9y}$.
Also, slope of line $5x+2y-10=0$ is $\frac{-5}{2}$
$\therefore$ Line is perpendicular to the tangent. So product of slopes $=-1$
$\therefore \frac{4x}{9y}\times \left(-\frac{5}{2}\right)=-1$
$\Rightarrow y=\frac{10x}{9}\cdots \left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$, we get
$4x^2-\frac{100x^2}{9}=36$
$\Rightarrow -64x^2=324$
which gives imaginary $x$.
Hence, there is no point on the curve at which tangent is perpendicular to the given line.