For the hyperbola `x^2/100-y^2/25=1`, prove that `SA*S'A=25`, where S and S' are the foci and A is the vertex.

The foci of the hyperbola 9x^2-16 y^2=144 are

Find the area of ellipse x^2/6+y^2/25=1 .

Find the equation of the hyperbola in the form x^2/a^2-y^2/b^2=1 given that : the distance between foci is 10 and LR=9/2

Find the equation of the hyperbola in the form x^2/a^2-y^2/b^2=1 given that : distance between the foci is 12 and e=3

If a point P on the ellipse x^2/a^2+y^2/b^2=1 is (a cos theta, b sin theta) and S, S' are the foci, then prove that SP*S'P=a^2sin^2theta+b^2cos^2theta .

The line segment joining the foci of the hyperbola x^2 – y^2 +1 = 0 is one of thediameters of a circle. The equation of the circle is

P is any point on the ellipse x^2/144+y^2/36=1 and S ans S' are foci, then the perimeter of triangle SPS' is :

The eccentricity of the hyperbola x ^(2) - y^(2) =25 is