The parabola `y^2 = 4` ax goes through the point (3,-2). Obtain the length of latus-rectum and the coordiante of the focus.

If the parabola y^2=4a x\ passes through the point (3,2) then find the length of its latus rectum.

The focus of the parabola y^2= 4ax is :

Find the length of latus rectum of the ellipse. x^2/16+y^2/9=1 .

L L ' is the latus rectum of the parabola y^2 = 4ax and PP' is a double ordinate drawn between the vertex and the latus rectum. Show that the area of the trapezium P P^(prime)L L ' is maximum when the distance P P ' from the vertex is a//9.

Find the length of latus rectum of the ellipse x^(2)/25+y^(2)/16=1 .

The length of the latus-rectum of the parabola x^2- 4x-8y +12= 0 is:

The directrix of the parabola y^2=4ax is :

Find the vertex and length of latus rectum of the parabola x^(2)=-4(y-a) .