Find the equation for the ellipse that satisfies the given conditions: Length of major axis $26$ foci $(±5,\,0)$
Length of major axis $=26 ;$ foci $=(\pm 5,\,0)$
since the foci are on the $x-$ axis, the major axis is along the $x-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ where a is the semimajor axis.
Accordingly, $2 a=26 \Rightarrow a=13$ and $c=5$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore 13^{2}=b^{2}+5^{2}$
$\Rightarrow 169=b^{2}+25$
$\Rightarrow b^{2}=169-25$
$\Rightarrow b=\sqrt{144}=12$
Thus, the equation of the ellipse is $\frac{x^{2}}{13^{2}}+\frac{y^{2}}{12^{2}}=1$ or $\frac{x^{2}}{169}+\frac{y^{2}}{144}=1$
$P$ is a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with $AA'$ as the major axis. Then the maximum value of the area of $\Delta APA'$ is
Let $S$ and $S\,'$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S\,'BS$ is a right angled triangle with right angle at $B$ and area $(\Delta S\,'BS) = 8\,sq.$ units, then the length of a latus rectum of the ellipse is
Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$ | $List-II$ |
If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
The equation of tangent and normal at point $(3, -2)$ of ellipse $4{x^2} + 9{y^2} = 36$ are
Let $E$ be the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ and $C$ be the circle ${x^2} + {y^2} = 9$. Let $P$ and $Q$ be the points $(1, 2)$ and $(2, 1)$ respectively. Then