# Locus of points of intersection of normals of two points on parabola $y=x^2$?

## 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?

• 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? – amd May 22 '18 at 6:30
• @amd I meant latus rectum, I have edited the question. – Hema May 22 '18 at 6:40
• @amd thanks I got it! – Hema May 22 '18 at 6:52
• 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? – amd May 22 '18 at 6:52
• @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) – Hema May 22 '18 at 6:58

## 1 Answer

Equation of first normal:

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

Equation of second normal:

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

$(2)-(1)$,

\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}