Parabola and locus....

Question

..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.

Answer 1

Hint: Consider the following points:

$1$. 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.

Answer 2

The points of intersection of tangents at points with parameters $t_1,t_2,t_3$ on $y^2=4ax$ are $\displaystyle A(at_1t_2, a(t_1+t_2)),B(at_2t_3, a(t_2+t_3)), A(at_3t_1, a(t_3+t_1))$

$A,B$ lie on $x^2=4by$ and hence we get $(at_2^2)t_1^2-4bt_1-4bt_2=0$ and $(at_2^2)t_3^2-4bt_3-4bt_2=0$ and hence $t_1,t_3$ are roots of $(at_2^2)z^2-4bz-4bt_2=0$

Hence $t_1+t_3 = \dfrac{4b}{at_2^2}, t_1+t_3=\dfrac{4b}{at_2^2}$

From this we can easily show that $C$ also lies on $x^2=4by$