Find the equation for the ellipse that satisf | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) 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

Vedclass · Accepted Answer

$\frac{x^{2}}{64} + \frac{y^{2}}{100} = 1$

Vedclass · Answer

(A) 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 semi-major axis.Accordingly,$2b = 16 \Rightarrow b = 8$ and $c = 6$.It is known that $a^{2} = b^{2} + c^{2}$.$	herefore 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 $\frac{x^{2}}{64} + \frac{y^{2}}{100} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$,foci $(0, \pm 6)$.

Similar Questions

The equation of the ellipse passing through the origin and having foci at $(1, 0)$ and $(3, 0)$ is .....

Tangents are drawn to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ at all the ends of its latus rectum. The area of the quadrilateral so formed (in sq. units) is

If the maximum distance of the normal to the ellipse $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$,where $b < 2$,from the origin is $1$,then the eccentricity of the ellipse is:

The number of real tangents that can be drawn to the ellipse $3x^2 + 5y^2 = 32$ passing through $(3, 5)$ is

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module