The length of the latus rectum of an ellipse | vedclass

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

Question

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 6$,which implies $b^

Vedclass · Accepted Answer

$25 e^2-40 e+16=0$

Vedclass · Answer

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 6$,which implies $b^2 = 3a$. The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a - ae = a(1 - e) = \frac{5}{3}$. We know $b^2 = a^2(1 - e^2)$. Substituting $b^2 = 3a$,we get $3a = a^2(1 - e^2)$,so $3 = a(1 - e^2) = a(1 - e)(1 + e)$. Since $a(1 - e) = \frac{5}{3}$,we have $3 = \frac{5}{3}(1 + e)$. This gives $1 + e = \frac{9}{5}$,so $e = \frac{9}{5} - 1 = \frac{4}{5}$. Now,we check which equation is satisfied by $e = \frac{4}{5}$. For option $A$: $25(\frac{4}{5})^2 - 40(\frac{4}{5}) + 16 = 25(\frac{16}{25}) - 32 + 16 = 16 - 32 + 16 = 0$. Thus,$e$ satisfies $25e^2 - 40e + 16 = 0$.

The length of the latus rectum of an ellipse is $6$ units and the distance between a focus and its nearest vertex on the major axis is $\frac{5}{3}$ units. If $e$ is the eccentricity of this ellipse,then $e$ satisfies the equation:

Similar Questions

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area,its eccentricity is

Let the equations of two ellipses be $E_1: \frac{x^2}{3} + \frac{y^2}{2} = 1$ and $E_2: \frac{x^2}{16} + \frac{y^2}{b^2} = 1$. If the product of their eccentricities is $\frac{1}{2}$,then the length of the minor axis of ellipse $E_2$ is

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

Let $A = \{(x, y) : y = mx + 1\}$,$B = \{(x, y) : x^2 + 4y^2 = 1\}$,and $C = \{(\alpha, \beta) : (\alpha, \beta) \in A \text{ and } (\alpha, \beta) \in B \text{ and } \alpha > 0\}$. If set $C$ is a singleton set,then the sum of all possible values of $m$ is:

The eccentric angle of the end point of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module