Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola $16y^2 - 9x^2 = 1 $ is

Find the equation of the hyperbola satisfying the give conditions : Vertices $(\pm 7,\,0)$,  $e=\frac{4}{3}$

Let the tangent to the parabola $y^2=12 x$ at the point $(3, \alpha)$ be perpendicular to the line $2 x+2 y=3$.Then the square of distance of the point $(6,-4)$from the normal to the hyperbola $\alpha^2 x^2-9 y^2=9 \alpha^2$at its point $(\alpha-1, \alpha+2)$ is equal to $........$.

The straight line $x + y = \sqrt 2 p$ will touch the hyperbola $4{x^2} - 9{y^2} = 36$, if

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

The locus of the point of intersection of the lines $(\sqrt{3}) kx + ky -4 \sqrt{3}=0$ and $\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is .............