# Locus of point of intersection of tangents

1. Sep 13, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
Locus of the point of intersection of tangents to the parabolas y$^{2}$=4(x+1) and y$^{2}$=8(x+2) which are at right angles, is

2. Relevant equations
Equation of tangent for first parabola
t$_{1}$y=x+1+at$_{1}$$^{2}$
Equation of tangent for second parabola
t$_{2}$y=x+2+bt$_{2}$$^{2}$

3. The attempt at a solution
Let us assume that the point of intersection of tangents is (h,k)
Since it lies on the tangent
∴kt$_{1}$=h+1+at$_{1}$$^{2}$
$\Rightarrow$at$_{1}$$^{2}$-kt$_{1}$+(h+1)=0
t$_{1}$t$_{2}$=h+1 (product of roots)
Also t$_{1}$t$_{2}$=-1 (Since they are at right angles)
∴required locus = x+2=0

2. Sep 13, 2012

### ehild

I can not follow your derivation. What have you got for a and b to make the equations tangent to the parabolas? The product of roots in the equation in red is (h+1)/a. t2 is not the other root of the equation for the first tangent line. Use the equations of both tangent lines to get the intersection.

I also got x=const. for the locus, but different from yours.

ehild

Last edited: Sep 13, 2012
3. Sep 14, 2012

### utkarshakash

Hey actually i arrived at the wrong answer because I only considered the first equation but not second. By solving both I get x+2=0 and x+4=0. Adding these two yields 2x+6=0 or x+3=0 which is the required locus.

P.S.- I wrote product of roots as h+1 because a=1. so (h+1)/a=(h+1)/1=(h+1)

4. Sep 14, 2012

### ehild

x+2=0 and x+4=0 contradict to each other, so that is not the right way to go, but the end result is correct. I got the same .

But you kept that a=1 top secret. It took me some time to find out what your equations mean.

ehild

5. Sep 15, 2012

### ehild

As it is a nice problem, and the OP has solved it, I write out the steps in a more ordered form.

We have two parabolas

(1)$$y^2=4(x+1)$$
(2)$$y^2=8(x+2)$$

The tangent lines are
$$x=ty+q$$.
Find q in terms of t so the line is tangent to the parabola, that is, the line and the parabola have a single common point. Substitute for x into the equation of parabola(1):
(1)
$$y^2-4t_1y-4(q_1+1)=0$$
The quadratic equation has a single solution, so the discriminant is zero--> $$D=16t_1^2+16(q_1+1)=0\rightarrow q_1=-(t_1^2+1)$$
(2)Substitute the equation of the tangent line for x in the equation of the second parabola: $$y^2-8t_21y-8(q_1+2)=0$$
The discriminant is $$D=64t_2^2+32(q_2+2)=0\rightarrow q_2=-2(t_1^2+1)$$

The tangent lines are perpendicular so $$t_1t_2=-1$$
Let t1=t, t2=-1/t. With that notation:
(1)$$x=ty-(t^2+1) \rightarrow y=x/t+(t+1/t)$$
(2)$$x=-y/t-2(-1/t^2+1) \rightarrow y=-tx-2(t+1/t)$$
Subtract the expressions for y to get the abscissa of the locus.

ehild

6. Sep 15, 2012

### utkarshakash

Thanks for helping.