Question

Let $y=f\left(x\right)$ be a parabola having $(0,\frac{3}{2})$ and $(0,0)$ as vertex and focus respectively. Then the number of roots of the equation $\frac{1}{2f\left(x\right)-3}-\frac{1}{2f\left(x\right)+3}+\frac{6}{x^2}=0$ is

Answer 1

Parabola having $(0,\frac{3}{2})$ and $(0,0)$ as vertex and focus respectively is
$x^2=-4\times \frac{3}{2}(y-\frac{3}{2})y=f\left(x\right)=-\frac{x^2}{6}+\frac{3}{2}\dots \left(1\right)$

Now, using equation $\left(1\right),$
$\frac{1}{2f\left(x\right)-3}-\frac{1}{2f\left(x\right)+3}+\frac{6}{x^2}=0\Rightarrow \frac{-3}{x^2}+\frac{3}{x^2-18}+\frac{6}{x^2}=0\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$
Hence, two solutions are there.