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The tangents intercept a distance of $4c$ on the tangent at the vertex.

The third tangent at the vertex is the Y axis.

The point interception of tangents are $(0,4c)$ and $(0,-4c)$

Let them interestect at (h,k)

The equation of tangent to the parabola $$y=mx+\frac am$$ $$\pm 4c=0+\frac am$$ $$m=\frac {\pm a}{4c}$$

The slope of the first tangent is $$\frac{4c-k}{0-h}=\frac {a}{4c}$$ $$16c^2-4ck=-ah$$ $$ax-4cy+16c^2=0$$

But the answer given is $y^2-4ax=16c^2$

I know I have considered the other equation yet. I have it with me, but I know how to apply it.

Aditya
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    It seems you're misreading the question statement: The tangents intercept a distance of 4c. It is the distance between the two intercepts that is constant $4c$, while the points can move. If it were meant to be like you formulated $(0, \pm 4c)$ , then the tangents are uniquely determined and there's no such thing as the locus. – Lee David Chung Lin Feb 03 '20 at 15:20
  • @LeeDavidChungLin that may be so. In that, could you please explain how I should solve it, because I have absolutely no idea. – Aditya Feb 03 '20 at 15:37
  • BTW, the correct answer should be $y^2-4ax=16 c^2$ and not $y^2-4ax=8c^2$. Don't know where you got that. – Lee David Chung Lin Feb 03 '20 at 16:45
  • @LeeDavidChungLin I misread it. $16c^2$ is correct – Aditya Feb 03 '20 at 16:48

1 Answers1

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Let the point of intersection be $P(h,k)$

which will satisfy $$y=mx+\dfrac am$$ i.e. $$k=mh+\dfrac am\implies m^2h-mk+a=0$$

if the two roots are $m_1,m_2$

$$m_1+m_2=\dfrac kh,m_1m_2=\dfrac ah\ \ \ \ (1)$$

Now the equation of the tangent through vertex $(0,0)$ is $x=0$ So, $y_k=\dfrac a{m_k}, k=1,2$

$$|4c|=|y_1-y_2|$$

$$\implies(4c)^2=a^2\dfrac{(m_1+m_2)^2-4m_1m_2}{(m_1m_2)^2}$$

Use $(1)$ to eliminate $m_1,m_2$

lab bhattacharjee
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