A parabola of latus rectum `l` touches a fixed equal parabola. The axes of two parabolas are parallel. Then find the locus of the vertex of the moving parabola.

Explanation

Let the fixed parabola be
`y^(2)=4ax` (1)

Since moving parabola and given parabola are equal, they have same latus rectum length.
So, latus rectum length.
`(y-k)^(2)=-4a(x-h)` (2)
On solving equation (1) and (2), we get
`(y-k)^(2)=-4a((y^(2))/(4a)-h)`
`rArr" "y^(2)-2ky+k^(2)=-y^(2)+4ah`
`rArr" "2y^(2)-2ky+k^(2)-4ah=0`
Since the two parabolas touch each other, above equation has two equal roots and therefore, discriminant is zero.
`rArr" "4k^(2)-8(k^(2)-4ah)=0`
`rArr" "-4k^(2)+32ah=0`
`rArr" "y^(2)=8ax`, which is locus of the vertex of the moving parabola.
Thus locus is parabola.