Find the equation of the ellipse, whose length of the major axis is $20$ and foci are $(0,\,\pm 5)$
since the foci are on $y-$ axis, the major axis is along the $y-$ axis. So, equation of the cllipse is of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$
Given that
$a=$ semi-major axis $=\frac{20}{2}=10$
and the relation $c^{2}=a^{2}-b^{2}$ gives
$5^{2}=10^{2}-b^{2} $ i.e., $b^{2}=75$
Therefore, the equation of the ellipse is
$\frac{x^{2}}{75}+\frac{y^{2}}{100}=1$
The locus of mid points of parts in between axes and tangents of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ will be
Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P, Q$ and $Q^{\prime}$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M \left( P , Q ^{\prime}\right)$ be the mid-point of the line segment joining $P$ and $Q ^{\prime}$. Then the maximum possible value of the distance between $M ( P , Q )$ and $M \left( P , Q ^{\prime}\right)$, as $P, Q$ and $Q^{\prime}$ vary on $E$, is. . . . .
The product of the lengths of perpendiculars from the foci on any tangent to the ellipse $3x^2 + 5y^2 = 1$, is
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.
If $PQ$ is a double ordinate of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies