An ellipse passing through (4 sqrt{2}, 2 sqrt | vedclass

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

Question

(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \ Given $ae = 4$. \ Let the equation of the ellipse be $\frac{x^2}{a

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \ Given $ae = 4$. \ Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. \ Since $b^2 = a^2(1 - e^2) = a^2 - (ae)^2 = a^2 - 16$,we substitute the point $(4\sqrt{2}, 2\sqrt{6})$: \ $\frac{(4\sqrt{2})^2}{a^2} + \frac{(2\sqrt{6})^2}{a^2 - 16} = 1$ \ $\frac{32}{a^2} + \frac{24}{a^2 - 16} = 1$ \ $32(a^2 - 16) + 24a^2 = a^2(a^2 - 16)$ \ $32a^2 - 512 + 24a^2 = a^4 - 16a^2$ \ $a^4 - 72a^2 + 512 = 0$ \ $(a^2 - 64)(a^2 - 8) = 0$ \ Since $a > ae$,$a^2$ must be greater than $16$,so $a^2 = 64$ and $a = 8$. \ Using $ae = 4$,we get $8e = 4$,which implies $e = \frac{1}{2}$.

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

Similar Questions

The lines $y=2x+\sqrt{76}$ and $2y+x=8$ touch the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$. If the point of intersection of these two lines lies on a circle whose centre coincides with the centre of that ellipse,then the equation of that circle is

The eccentricity of the ellipse $4x^2 + 25y^2 = 100$ is

If $(1, -2)$ is the focus and $x+y-2=0$ is the directrix of the ellipse $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$,then its eccentricity is

If the eccentric angles of the extremities of a focal chord (other than the major axis) of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ are $\alpha$ and $\beta$,then $\frac{\cot(\alpha/2)}{\tan(\beta/2)}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module