The vertex of ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ are :

The foci of the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$ are :

The maximum distance of the centre of the ellipse $\frac{x^2}{81}+\frac{y^2}{25}=1$ from a normal to the ellipse is

If the midpoint of the chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$is $(0,3)$

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}$ =1 and $e_2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e_1{e}_2=1$ then equation of the hyperbola is

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1\text{and}{e}_2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e_1{e}_2=1,$ then the equation of the hyperbola is

Find the equations of tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ at a point whose abscissa is $2\sqrt{2}$ and ordinate is positive

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then find the value of $b^2$

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide write the value of $b^2\stackrel{.}{}$

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide write the value of $b^2\stackrel{.}{}$

If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{125}$ coincide, the find the value of $b^2$.