Question

If the focus of the parabola $x^2-ky+3=0$ is $(0,2)$, then the value(s) of $k$ is/are

• A. $4$
• B. $6$
• C. $3$
• D. $2$

Answer 1

Answer: (B), (D)

$2$

Given parabola is
$x^2-ky+3=0$
or $x^2=k(y-\frac{3}{k})$
Le $x=X,y-\frac{3}{k}=Y$
then the parabola is
$X^2=kY$
whose focus is $(0,\frac{k}{4})$.
Therefore, the focus of $x^2=k(y-\frac{3}{k})$ is
$(0,\frac{3}{k}+\frac{k}{4})=(0,2)$
$\therefore \frac{3}{k}+\frac{k}{4}=2$
$\Rightarrow 12+k^2=8k$
$\Rightarrow k^2-8k+12=0$
$\Rightarrow (k-6)(k-2)=0$
$\Rightarrow k=2,6$