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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.

Brahmagupta
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1 Answers1

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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. $a^2\gt 8b^2$.

As harry pointed out in the comments, one has to consider the case where one of the tangents is the $y$-axis.

Two tangents drawn to the parabola $y^2=4ax$ from $(0,k)$ where $k\not=0$ are $x=0$ and $y=\frac akx+k$. It follows from $u^2=4bv,\frac ak=-\frac{2a}{u}$ and $k=2b+v$ that $bk^2-a^2k+2a^2b=0$. So, one has to have $a^2\geqslant 8b^2$. If $a^2=8b^2$, then two tangents drawn to the parabola $y^2=4ax$ from a point $(0,4b)$ are $x=0$ and $ax-4by+16b^2=0$ (at $(2a,8b)$) which are normals to the parabola $x^2=4by$ (at $(-a,2b)$).

In conclusion, the answer is $a^2\geqslant 8b^2$.

mathlove
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  • Great answer! But the forms for the tangent and normal you've used can't be used to represent another, single line that also satisfies the condition; the y axis. If I'm not wrong, the answer would then be $a^2 \geq 8b^2$, since the minimum condition for such a point as described in the question to exist is satisfied by any one line that we'd get if $a^2=8b^2$, since then we'd get a point where that line intersects the y axis. For the greater-than condition, we'd get three points. Could you verify if this is correct? – harry Sep 08 '21 at 02:33
  • @harry : Let us consider the case where $a=2\sqrt 2$ and $b=1$ satisfying $a^2=8b^2$. Two tangents drawn to the parabola $y^2=8\sqrt 2\ x$ from a point $(0,1)$ are $x=0$ and $y=2\sqrt 2\ x+1$. However, the line $y=2\sqrt 2\ x+1$ is not a normal to the parabola $x^2=4y$. – mathlove Sep 08 '21 at 12:00
  • If $a=2\sqrt{2}$ and $b=1$, then the equation for the slope would be $2m^2-2\sqrt{2}m+1=0$, the singular solution to which is $m=\frac{1}{\sqrt{2}}$. Going by that, the obtained line $x-2y+4\sqrt{2}=0$ is tangent to the horizontal parabola and normal to the vertical parabola as shown here. I checked the angles, and they seem to work out. Could you clarify if the first comment was correct? – harry Sep 08 '21 at 19:48
  • I don't think drawing tangents from any point would suffice; the point has properties, which the slope of the tangents, a and b determine since we get the point on solving equations involving them, but the question only asks for the condition for such points to exist, which seems to be the greater-than-or-equal-to inequality. – harry Sep 08 '21 at 19:52
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    @harry : Thanks, you are right. You have a typo in $x−2y+4\sqrt 2=0$. – mathlove Sep 09 '21 at 07:44