Equation of locus of two tangents

**arze** · October 13th 2009, 10:47 PM

If the normal at $P(ap^2,2ap)$ to the parabola $y^2=4ax$ meets the curve again at $Q(aq^2,2aq)$ prove that $p^2+pq+2=0$ . Prove that the equation of the locus of the point of intersection of he tangents to the parabola at P and Q is
$y^2(x+2a)+4a^3=0$ .

I've done the first part of proving $p^2+pq+2=0$ .
for this second part, I tried to find a parametric equation for the locus the point of intersection of the tangents. Found the equation of the two tangents, then the point of intersection. The problem I have is there is no p or q in the equation I'm supposed to prove, while the point is in terms of p and q. Need someone to point me in the right direction and I should be able to finish off. Thanks!

**Opalg** · October 14th 2009, 12:46 PM

Originally Posted by arze

If the normal at $P(ap^2,2ap)$ to the parabola $y^2=4ax$ meets the curve again at $Q(aq^2,2aq)$ prove that $p^2+pq+2=0$ . Prove that the equation of the locus of the point of intersection of he tangents to the parabola at P and Q is
$y^2(x+2a)+4a^3=0$ .

I've done the first part of proving $p^2+pq+2=0$ .
for this second part, I tried to find a parametric equation for the locus the point of intersection of the tangents. Found the equation of the two tangents, then the point of intersection. The problem I have is there is no p or q in the equation I'm supposed to prove, while the point is in terms of p and q. Need someone to point me in the right direction and I should be able to finish off. Thanks!

You should find that the point of intersection is P = (apq,a(p+q)). But you know that $p^2+pq+2=0$ . Solve that for q, and substitute that value for q into the coordinates of P. That gives $P = (-ap^2-2a,-2a/p)$ . So if P = (x,y) then $x = -ap^2-2a$ and $y = -2a/p$ . Now eliminate p from those equations for x and y.

**arze** · October 14th 2009, 04:34 PM

Originally Posted by Opalg

You should find that the point of intersection is P = (apq,a(p+q)). But you know that $p^2+pq+2=0$ . Solve that for q, and substitute that value for q into the coordinates of P. That gives $P = (-ap^2-2a,-2a/p)$ . So if P = (x,y) then $x = -ap^2-2a$ and $y = -2a/p$ . Now eliminate p from those equations for x and y.

Thanks! But how to eliminate p? Can you explain abit more? thanks

**mr fantastic** · October 14th 2009, 06:30 PM

Originally Posted by arze

Thanks! But how to eliminate p? Can you explain abit more? thanks

Make p the subject in the parametric equation for y and substitute the resulting expression into the parametric equation for x.

Thread: Equation of locus of two tangents

LinkBack

Thread Tools

Display

Equation of locus of two tangents

Similar Math Help Forum Discussions

locus of tangents

Ellipse and locus of point of intersection of two tangents

Equation Of A Locus

equation of locus

Equation of Locus

Search Tags