# Parabola and locus....

..If a triangle is formed by any three tangents of the parabola $y^2=4ax$, two ofwhose vertices lie on the parabola $x^2=4by$, then find the locus of the third vertex.

Please somebody. Help I could not solve it.

• $y^2=4a$?? not $y^2=4ax$??
– user371838
Jan 22 '17 at 6:42
• Jan 22 '17 at 7:06

$1$. We know that the parametrization of a point on a parabola $y^2=4ax$ is $(at^2,2at)$.
$2$. We know that the slope of the tangent at this point is $\frac {1}{t}$.
$3$. Find the equation of all three tangents and find their intersection points.
$4$. Consider any two points to lie on the parabola $x^2=4by$ and the intersection of the remaining pair is the third point.
$5$. The locus of the third vertex can be found by eliminating the three parameters from four equations we get ( two by substituting twould points in $x^2=4by$ and the other two are the x and y coordinates of the third point).
I get the answer to be $\boxed{x^2=4by}$. Hope it helps.