This is a question I found: P $(t_1,t_1^2) $ and Q $(t_2,t_2^2)$ are two points on a parabola such that PQ subtends a right angle at the vertex. What is the locus of points of intersection of normals at P and Q? The answer given is $x^2 = 2(2y-3)$.

Now I found that $ t_1t_2 = -1$. The tangents at P and Q will be given by the equations $$ y = 2xt_1 - t_1^2 $$ and $$ y = 2xt_2 - t_2^2 $$. I concluded that the normals will have the negative reciprocal as slope and are given by $$ y = -x/2t_1 +c1 $$ and $$ y = -x/2t_2 +c2 $$. On substituting $(t_1,t_1^2) $ in first normal equation I get $c_1 = (2x^2t_1+1/2t_1)$. This doesn't seem to be right. Is my method right? Also, how do I proceed to find the locus of points of intersection of these normals?

  • 1
    $\begingroup$ You need to substitute for both $x$ and $y$. That aside, how is the given answer $x^2=2(2y-3)$ the length of anything? $\endgroup$ – amd May 22 '18 at 6:30
  • $\begingroup$ @amd I meant latus rectum, I have edited the question. $\endgroup$ – Hema May 22 '18 at 6:40
  • $\begingroup$ @amd thanks I got it! $\endgroup$ – Hema May 22 '18 at 6:52
  • $\begingroup$ The question as written makes even less sense now. The latus rectum is a line segment, so what is the locus of a family of line segments? $\endgroup$ – amd May 22 '18 at 6:52
  • $\begingroup$ @amd I think it is asking to find the length of the latus rectum of locus of points of intersection.. my solution gives these lines: x = -2 t1 t2(t1+t2) and y = {(t1+t2)^2-t1t2}+(1/2), the next line is that required locus is x^2 = 2(2y-3) $\endgroup$ – Hema May 22 '18 at 6:58

Equation of first normal:

$$y-t^2=-\frac{1}{2t}(x-t) \tag{1}$$

Equation of second normal:

$$y-\frac{1}{t^2}=\frac{t}{2}\left( x+\frac{1}{t} \right) \tag{2}$$


\begin{align} \left( t^2-\frac{1}{t^2} \right) &= \left( \frac{t}{2}+\frac{1}{2t} \right) x \\ x &= 2 \left( t-\frac{1}{t} \right) \\ y &= t^2+\frac{1}{t^2}-\frac{1}{2} \\ &= \left( t-\frac{1}{t} \right)^2+\frac{3}{2} \\ &= \frac{x^2}{4}+\frac{3}{2} \\ x^2 &= 2(2y-3) \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.