Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
since denominator of $\frac{x^{2}}{25}$ is larger than the denominator of $\frac{y^{2}}{9},$ the major axis is along the $x-$ axis.
Comparing the given equation with $\frac{x^{2}}{a^{2}}$ $+\frac{y^{2}}{b^{2}}$ $=1,$ we get $a=5$ and $b=3$ . Also $c=\sqrt{a^{2}-b^{2}}=\sqrt{25-9}=4$
Therefore, the coordinates of the foci are $(-4,\,0)$ and $(4,\,0),$ vertices are $(-5,\,0)$ and $(5,\,0).$ Length of the major axis is $10$ units length of the minor axis $2b$ is $6$ units and the eccentricity is $\frac{4}{5}$ and latus rectum is $\frac{2 b^{2}}{a}=\frac{18}{5}$.
If the length of the latus rectum of the ellipse $x^{2}+$ $4 y^{2}+2 x+8 y-\lambda=0$ is $4$ , and $l$ is the length of its major axis, then $\lambda+l$ is equal to$......$
If the curves, $\frac{x^{2}}{a}+\frac{y^{2}}{b}=1$ and $\frac{x^{2}}{c}+\frac{y^{2}}{d}=1$ intersect each other at an angle of $90^{\circ},$ then which of the following relations is TRUE ?
The sum of the focal distances of any point on the conic $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$ is
The normal at a variable point $P$ on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}= 1$ of eccentricity e meets the axes of the ellipse in $ Q$ and $R$ then the locus of the mid-point of $QR$ is a conic with an eccentricity $e' $ such that :
Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $=\frac{x^2}{36}+\frac{y^2}{16}=1$.If $(1, \alpha)$ lies on $C$, then $10 \alpha^2$ is equal to $.........$