# Equation of locus of two tangents

• Oct 13th 2009, 10:47 PM
arze
Equation of locus of two tangents
If the normal at \$\displaystyle P(ap^2,2ap)\$ to the parabola \$\displaystyle y^2=4ax\$ meets the curve again at \$\displaystyle Q(aq^2,2aq)\$ prove that \$\displaystyle 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
\$\displaystyle y^2(x+2a)+4a^3=0\$.

I've done the first part of proving \$\displaystyle 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!
• Oct 14th 2009, 12:46 PM
Opalg
Quote:

Originally Posted by arze
If the normal at \$\displaystyle P(ap^2,2ap)\$ to the parabola \$\displaystyle y^2=4ax\$ meets the curve again at \$\displaystyle Q(aq^2,2aq)\$ prove that \$\displaystyle 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
\$\displaystyle y^2(x+2a)+4a^3=0\$.

I've done the first part of proving \$\displaystyle 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 \$\displaystyle p^2+pq+2=0\$. Solve that for q, and substitute that value for q into the coordinates of P. That gives \$\displaystyle P = (-ap^2-2a,-2a/p)\$. So if P = (x,y) then \$\displaystyle x = -ap^2-2a\$ and \$\displaystyle y = -2a/p\$. Now eliminate p from those equations for x and y.
• Oct 14th 2009, 04:34 PM
arze
Quote:

Originally Posted by Opalg
You should find that the point of intersection is P = (apq,a(p+q)). But you know that \$\displaystyle p^2+pq+2=0\$. Solve that for q, and substitute that value for q into the coordinates of P. That gives \$\displaystyle P = (-ap^2-2a,-2a/p)\$. So if P = (x,y) then \$\displaystyle x = -ap^2-2a\$ and \$\displaystyle 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
• Oct 14th 2009, 06:30 PM
mr fantastic
Quote:

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.