The lines $y=2x+\sqrt{76}$ and $2y+x=8$ touch the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$. If the point of intersection of these two lines lies on a circle whose centre coincides with the centre of that ellipse,then the equation of that circle is

$A$ vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $PQR$,$\Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$,then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

The eccentricity of the conic $36x^2 + 144y^2 - 36x - 96y - 119 = 0$ is

The equation of the ellipse passing through the origin and having foci at $(1, 0)$ and $(3, 0)$ is .....

$A$ point moves such that the sum of its distances from two fixed points $(ae, 0)$ and $(-ae, 0)$ is always $2a$. Then the equation of its locus is