The eccentricity of an ellipse whose length o | vedclass

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

Question

(A) The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$ and the distance between its foci is $2ae$.Given that $\frac{2b^2}{a} = 2ae$,we h

Vedclass · Accepted Answer

$2 \sin 18^{\circ}$

Vedclass · Answer

(A) The length of the latus rectum of an ellipse is $\frac{2b^2}{a}$ and the distance between its foci is $2ae$.Given that $\frac{2b^2}{a} = 2ae$,we have $b^2 = a^2e$.Using the relation $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2e$.Dividing by $a^2$,we obtain $1 - e^2 = e$,which simplifies to $e^2 + e - 1 = 0$.Solving this quadratic equation for $e$ (where $e > 0$),we get $e = \frac{-1 + \sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}$.Since $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$,it follows that $e = 2 \sin 18^{\circ}$.

The eccentricity of an ellipse whose length of latus rectum is equal to the distance between its foci is

Similar Questions

The point $(4, -3)$ with respect to the ellipse $4x^2 + 5y^2 = 1$ is:

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

$A$ tangent having slope $m = -1/2$ to the ellipse $3x^2 + 4y^2 = 12$ intersects the $X$-axis and $Y$-axis at the points $A$ and $B$ respectively. If $O$ is the origin, then the area of $\Delta AOB$ is ...

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

If $P$ is any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $S, S^{\prime}$ are its foci,then the maximum area (in sq. units) of $\Delta SPS^{\prime} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module