Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$ foci $(0,\,±6)$
Length of minor axis $=16 ;$ foci $=(0,\,\pm 6)$
since the foci are on the $y-$ axis, the major axis is along the $y-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,$ where a is the semimajor axis.
Accordingly, $2 b=16 \Rightarrow b=8$ and $c=6$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore a^{2}=8^{2}+6^{2}=64+36=100$
$\Rightarrow a=\sqrt{100}=10$
Thus, the equation of the ellipse is $\frac{x^{2}}{8^{2}}+\frac{y^{2}}{10^{2}}=1$ or $\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$
If $C$ is the centre of the ellipse $9x^2 + 16y^2$ = $144$ and $S$ is one focus. The ratio of $CS$ to major axis, is
The position of the point $(4, -3)$ with respect to the ellipse $2{x^2} + 5{y^2} = 20$ is
If tangents are drawn from point $P(3\ sin\theta + 4\ cos\theta , 3\ cos\theta\ -\ 4\ sin\theta)$ , $\theta = \frac {\pi}{8}$ to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$ then angle between the tangents is
Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)
$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$
If the points of intersection of two distinct conics $x^2+y^2=4 b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3 x^2$, then $3 \sqrt{3}$ times the area of the rectangle formed by the intersection points is............................