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..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.

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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.

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