Question

If the curves $3x^2+4y^2=12$ and $k{x}^2+2y^2=2k$ cut orthogonally then $k^2+3k=?$

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $4$

$3x^2+4y^2=12⟹y^2=\frac{12-3x^2}{4}\cdots \left(1\right)$

Put it in second equation

$⟹k{x}^2+\frac{12-3x^2}{2}=2k⟹x^2=\frac{4k-12}{2k-3}\cdots \left(2\right)$

Put it in $\left(1\right)$

$⟹y^2=\frac{3k}{2k-3}$

Differentiating the equations

$\frac{dy}{dx}$ of first curve is $\frac{-3x}{4y}$

$\frac{dy}{dx}$ of second curve is $\frac{-kx}{2y}$

According to Question,

$\frac{3k{x}^2}{8y^2}=-1\cdots \left(3\right)$

Put $\left(1\right)\left(2\right)$ in $\left(3\right)$

$3k\left(\frac{4k-12}{2k-3}\right)=-8\left(\frac{3k}{2k-3}\right)⟹k=1$

$⟹k^2+3k=4$