Let E_{1}: frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{ | vedclass

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

Question

(C) For ellipse $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e^{2} = 1 - \frac{b^{2}}{a^{2}}$.For ellipse $E_{2}$,the foci

Vedclass · Accepted Answer

$\frac{-1+\sqrt{5}}{2}$

Vedclass · Answer

(C) For ellipse $E_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e^{2} = 1 - \frac{b^{2}}{a^{2}}$.For ellipse $E_{2}$,the foci are $(0, b)$ and $(0, -b)$,so it is a vertical ellipse with $2c = 2b$,hence $c = b$. The vertices are at $(\pm a, 0)$,so the semi-major axis is $A = a$. The equation of $E_{2}$ is $\frac{x^{2}}{B^{2}} + \frac{y^{2}}{A^{2}} = 1$,where $A = a$. Since $c^{2} = A^{2} - B^{2}$,we have $b^{2} = a^{2} - B^{2}$,so $B^{2} = a^{2} - b^{2}$.The eccentricity of $E_{2}$ is $e^{2} = 1 - \frac{B^{2}}{A^{2}} = 1 - \frac{a^{2} - b^{2}}{a^{2}} = \frac{b^{2}}{a^{2}}$.Since both ellipses have the same eccentricity $e$,we have $e^{2} = \frac{b^{2}}{a^{2}}$.From the first equation,$e^{2} = 1 - e^{2}$,which implies $2e^{2} = 1$,so $e = \frac{1}{\sqrt{2}}$.Wait,re-evaluating the condition: The foci of $E_{2}$ are $(0, b)$ and $(0, -b)$,so $c_{2} = b$. The vertices of $E_{2}$ are $(\pm a, 0)$,so the semi-minor axis is $a$. The major axis is along the $y$-axis,so $A_{2} = c_{2}/e = b/e$. The relation $c_{2}^{2} = A_{2}^{2} - B_{2}^{2}$ gives $b^{2} = (b/e)^{2} - a^{2}$.Thus $a^{2} = \frac{b^{2}}{e^{2}} - b^{2} = b^{2}(\frac{1-e^{2}}{e^{2}})$.Since $1-e^{2} = b^{2}/a^{2}$,we have $a^{2} = b^{2}(\frac{b^{2}/a^{2}}{e^{2}}) = \frac{b^{4}}{a^{2}e^{2}}$,so $a^{4}e^{2} = b^{4}$,which means $a^{2}e = b^{2}$.Substituting $b^{2} = a^{2}e$ into $e^{2} = 1 - b^{2}/a^{2}$,we get $e^{2} = 1 - e$,or $e^{2} + e - 1 = 0$.Solving for $e > 0$,we get $e = \frac{-1 + \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 + \sqrt{5}}{2}$.

Similar Questions

Let a tangent drawn at any point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut the $X$-axis at $Q$. Let $R$ be the image of $Q$ with respect to $y=x$. If $S$ is a circle with $QR$ as its diameter,then the fixed point through which the circle $S$ passes is

If the distance between the foci of an ellipse is equal to the length of the latus rectum, then its eccentricity is

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$. Then the maximum perpendicular distance of $P$ from the line $x - y = 10$ is (in $\sqrt{2}$)

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a > b$) is twice the area of the ellipse,then the eccentricity of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module