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.

  • 1
    $\begingroup$ 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. $\endgroup$ Feb 3 '20 at 15:20
  • $\begingroup$ @LeeDavidChungLin that may be so. In that, could you please explain how I should solve it, because I have absolutely no idea. $\endgroup$
    – Aditya
    Feb 3 '20 at 15:37
  • $\begingroup$ 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. $\endgroup$ Feb 3 '20 at 16:45
  • $\begingroup$ @LeeDavidChungLin I misread it. $16c^2$ is correct $\endgroup$
    – Aditya
    Feb 3 '20 at 16:48

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, $y_k=\dfrac a{m_k}, k=1,2$



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

  • $\begingroup$ Isn’t the vertex (0,0)? Also why is $y_k=\frac{a}{m_k}$ $\endgroup$
    – Aditya
    Feb 3 '20 at 16:16
  • $\begingroup$ @Aditya, there are two possible values of $m$ each corresponds to one value of $y$ $\endgroup$ Feb 3 '20 at 16:22
  • $\begingroup$ But how did you arrive at the relation? $\endgroup$
    – Aditya
    Feb 3 '20 at 16:31

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