Let S = 0 be an ellipse whose vertices are th | vedclass

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

Question

(A) The ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$. The foci are $(\pm ae, 0)$ and the extremities of the minor axis are $(0,

Vedclass · Accepted Answer

$\sqrt{\frac{1 - 2e^2}{1 - e^2}}$

Vedclass · Answer

(A) The ellipse $E$ is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$. The foci are $(\pm ae, 0)$ and the extremities of the minor axis are $(0, \pm b)$.For the ellipse $S = 0$,the vertices are $(0, \pm b)$,so its minor axis lies on the $x$-axis and its major axis lies on the $y$-axis.The equation of $S = 0$ is $\frac{x^2}{b_1^2} + \frac{y^2}{b^2} = 1$ where $b_1$ is the semi-minor axis of $S$.Since $S = 0$ passes through the foci $(\pm ae, 0)$,we have $\frac{(ae)^2}{b_1^2} + \frac{0^2}{b^2} = 1$,which gives $b_1^2 = a^2e^2$.For an ellipse with major axis $b$ and semi-minor axis $b_1$,the eccentricity $e_1$ satisfies $b_1^2 = b^2(1 - e_1^2)$.Substituting $b_1^2 = a^2e^2$ and $b^2 = a^2(1 - e^2)$,we get $a^2e^2 = a^2(1 - e^2)(1 - e_1^2)$.$e^2 = (1 - e^2)(1 - e_1^2) \Rightarrow 1 - e_1^2 = \frac{e^2}{1 - e^2}$.$e_1^2 = 1 - \frac{e^2}{1 - e^2} = \frac{1 - 2e^2}{1 - e^2}$.Thus,$e_1 = \sqrt{\frac{1 - 2e^2}{1 - e^2}}$.

Let $S = 0$ be an ellipse whose vertices are the extremities of the minor axis of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. If $S = 0$ passes through the foci of $E$,then its eccentricity is (considering the eccentricity of $E$ as $e$).

Similar Questions

If the tangent at a point on the ellipse $\frac{x^2}{27} + \frac{y^2}{3} = 1$ meets the coordinate axes at $A$ and $B,$ and $O$ is the origin,then the minimum area (in sq. units) of the triangle $OAB$ is

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

If $\tan \theta_1 \cdot \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at:

Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.
Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.
Which of the above statement$(s)$ is (are) true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module