Question

If $(2,-3);(5,k)$ are conjugate w.r.t $\frac{x^2}{4}+\frac{y^2}{2}=1$, then $k=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

The given equation $\frac{x^2}{4}+\frac{y^2}{2}=1$ is of an ellipse, as the eccentricity of the curve is $\frac{1}{\sqrt{2}}$ which is less than 1.

Conjugate points: (2,-3), (5,k)

Polar of the curve should pass through both the points.

Polar of (2,-3) = $\frac{x\left(2\right)}{4}+\frac{y(-3)}{2}=1\Rightarrow \frac{x}{2}-\frac{3y}{2}=1$

Polar of (5,k) = $\frac{x\left(5\right)}{4}+\frac{y\left(k\right)}{2}=1\Rightarrow \frac{5x}{4}+\frac{ky}{2}=1$

And the point (2,-3) should also lie on the polar of (5,k):

$\therefore \frac{5\left(2\right)}{4}-\frac{\left(k\right)3}{2}=1$

$\therefore k=1$