The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :
$\frac{{\left( {{a^2}\,\, - \,\,{b^2}} \right)\,\,ab}}{{{a^2}\,\, + \,\,{b^2}}}$
$\frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{a^2} + {b^2}} \right)ab}}$
$\frac{{\left( {{a^2}\,\, - \,\,{b^2}} \right)}}{{ab\,\,\left( {{a^2}\,\, + \,\,{b^2}} \right)}}$
$\frac{{{a^2}\,\, + \,\,{b^2}}}{{\left( {{a^2}\,\, - \,\,{b^2}} \right)\,\,ab}}$
The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $P Q$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points
The ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the straight line $y = mx + c$ intersect in real points only if
Find the equation of the ellipse whose vertices are $(±13,\,0)$ and foci are $(±5,\,0)$.
If the point of intersections of the ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{ b ^{2}}=1$ and the circle $x ^{2}+ y ^{2}=4 b , b > 4$ lie on the curve $y^{2}=3 x^{2},$ then $b$ is equal to:
The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is