The distance of the point `(sqrt6sectheta,sqrt3tantheta)` on the hyperbola from the centre of the hyperbola is 3 if `theta` is

The distance of the point (sqrt(6)sec theta,sqrt(3)tan theta) on the hyperbola from the centre of the hyperbola is 3 if theta is

The distance of the point (sqrt6 cos theta , sqrt2 sin theta) on the ellipse from the centre of the ellipse is 2 if theta =

sqrt(2)sectheta+tantheta=1

If tantheta+sectheta=sqrt3, then: theta=

Solve : sqrt(2)sectheta+tantheta=1

In a rectangular hyperbola, prove that the product of the focal distances of a point on it is equal to the square of its distance from the centre of the hyperbola .

Solve : sectheta-1=(sqrt2-1)tantheta

tantheta+sectheta=sqrt(3)

The locus of the point (sectheta+tantheta,sectheta-tantheta) where theta is parameter, is