Question

The equation of the tangents to $2x^2-3y^2=36$ which are parallel to the straight line $x+2y-10=0$ are

• A. $x+2y=0$
• B. $2x^2-3y^2=36$
• C. $x+2y+\frac{\sqrt{288}}{15}=0$
• D. None of these

Answer 1

The correct option is D None of these

The slope of the tangent to the curve is given by,
$4x-6y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{2x}{3y}$
This slope of the given line is $-\frac{1}{2}$.

So, we must have $\frac{2x}{3y}=-\frac{1}{2}$
$\Rightarrow x=-\frac{3y}{4}$
Putting $x=-\frac{3y}{4}$ in $2x^2-3y^2=36$ we get
$2\left(\frac{9y^2}{16}\right)-3y^2=36$
$\Rightarrow y^2=-\frac{288}{15}$
This does not give real values
Hence, the required tangent does not exist.