Equation for ellipse that satisfies given conditions: Vertices (0,,pm 13), foci (0,,±5) .

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

medium

Find the equation for the ellipse that satisfies the given conditions: Vertices $(0,\,\pm 13),$ foci $(0,\,±5)$.

Option A

Option B

Option C

Option D

Solution

Vertices $(0,\,\pm 13),$ foci $(0,\,±5)$

Here, the vertices are on 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, $a=13$ 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 $\frac{x^{2}}{12^{2}}+\frac{y^{2}}{13^{2}}=1$ or $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$

Standard 11

Mathematics

Share

Similar Questions

The length of the latus rectum of the ellipse $5{x^2} + 9{y^2} = 45$ is

easy

The line $x =8$ is the directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0,4 \sqrt{3})$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to $……..$

hard

(JEE MAIN-2023)

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:

hard

(JEE MAIN-2021)

Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a ponit $\mathrm{P}$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta),$ then $\beta$ is equal to

hard

(JEE MAIN-2020)

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 $………$

hard

(JEE MAIN-2023)