# Find the locus of point of intersection of the two tangents to the parabola $y^2=4ax$

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.

• 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. Feb 3 '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. Feb 3 '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. Feb 3 '20 at 16:45
• @LeeDavidChungLin I misread it. $16c^2$ is correct 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, $$y_k=\dfrac a{m_k}, k=1,2$$

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

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

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

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