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Find the condition on $a$ and $b$ so that the two tangents drawn to the parabola $y^2=4ax$ from a point are normals to the parabola $x^2=4by.$

The required condition is $a^2>8b^2$.I dont know how to prove it.I tried.

Let $(h,k)$ be the point from where tangents are drawn to the parabola $y^2=4ax$.Let $m_1,m_2$ be the slopes of the tangents.Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$ and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.

As $y=m_1x+\frac{a}{m_1}$ and $y=m_1x-2bm_1-bm^3_1$ are the same lines.Therefore,$\frac{a}{m_1}=-2bm_1-bm^3_1$

$bm^4+2bm^2+a=0$

And I am stuck.I cannot get the desired condition.Please help me.

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Equations of tangents are $y=m_1x+\frac{a}{m_1}$ and $y=m_2x+\frac{a}{m_2}$

I think this is correct.

and equations of normal to the parabola $x^2=4by$ is $y=m_1x-2bm_1-bm^3_1$ and $y=m_2x-2bm_2-bm^3_2$.

I don't think this is correct.

Let $(s,t)$ be a point on $x^2=4by$, i.e. $s^2=4bt$. Since $2x=4by'\Rightarrow y'=\frac{x}{2b}$, the equation of normal to the parabola at $(s,t)$ will be $y-t=-\frac{2b}{s}(x-s)\iff y=-\frac{2b}{s}x+2b+t$. Let $k=-\frac{2b}{s}$. Then, since $s=-\frac{2b}{k}$, the equation of the normal can be written as $$y=kx+2b+\frac{s^2}{4b}=kx+2b+\frac{1}{4b}\left(-\frac{2b}{k}\right)^2=kx+2b+\frac{b}{k^2}.$$

So, in our case, we have $$y=m_1x+2b+\frac{b}{m_1^2}\qquad\text{and}\qquad y=m_2x+2b+\frac{b}{m_2^2}.$$

Thus, since $y=m_1x+\frac{a}{m_1}$ and $y=m_1x+2b+\frac{b}{m_1^2}$ are the same lines, $$\frac{a}{m_1}=2b+\frac{b}{m_1^2}\iff 2bm_1^2-am_1+b=0.$$

Now, the discriminant is positive, so $(-a)^2-4\cdot 2b\cdot b\gt 0$, i.e. $a^2\gt 8b^2$.

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