Let an ellipse frac{x^2}{a^2} + frac{y^2}{b^2 | vedclass

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

Question

(D) Given the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a < b$. Since $a < b$, the eccentricity $e$ is given by $e^2 = 1 - \frac{

Vedclass · Accepted Answer

$\frac{8\sqrt{5}}{3}$

Vedclass · Answer

(D) Given the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a Since $a Given $e = \frac{\sqrt{5}}{3}$, so $e^2 = \frac{5}{9}$. Thus, $1 - \frac{a^2}{b^2} = \frac{5}{9} \implies \frac{a^2}{b^2} = 1 - \frac{5}{9} = \frac{4}{9}$. Let $a^2 = 4k$ and $b^2 = 9k$ for some constant $k > 0$. The ellipse passes through $(4, 3)$, so $\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1$. Substituting the values, $\frac{16}{4k} + \frac{9}{9k} = 1 \implies \frac{4}{k} + \frac{1}{k} = 1 \implies \frac{5}{k} = 1 \implies k = 5$. Therefore, $a^2 = 4(5) = 20$ and $b^2 = 9(5) = 45$. This gives $a = \sqrt{20} = 2\sqrt{5}$ and $b = \sqrt{45} = 3\sqrt{5}$. The length of the latus rectum for an ellipse with $a Length $= \frac{2(20)}{3\sqrt{5}} = \frac{40}{3\sqrt{5}} = \frac{40\sqrt{5}}{3 	imes 5} = \frac{8\sqrt{5}}{3}$.

Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a < b$, pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is:

Similar Questions

The number of values of $c$ such that the line $y = cx + c$,where $c \in R$,touches the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$ is:

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

The product of the lengths of perpendiculars from the foci on any tangent to the ellipse $3x^2 + 5y^2 = 1$ is:

If $P(x, y)$,$F_1 = (3, 0)$,$F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2 = \dots$

The eccentricity of the ellipse $9x^2 + 25y^2 = 225$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module