If S and S^{prime} are the foci of the ellips | vedclass

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

Question

(A) Given the equation of the ellipse: $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here,$a^2=25 \implies a=5$ and $b^2=16 \implies b=4$. The eccentricity $e$ i

Vedclass · Accepted Answer

$4+S^{\prime}P$

Vedclass · Answer

(A) Given the equation of the ellipse: $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here,$a^2=25 \implies a=5$ and $b^2=16 \implies b=4$. The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. By the definition of an ellipse,the sum of focal distances of any point $P$ on the ellipse is $PS+PS^{\prime} = 2a$. Thus,$PS+PS^{\prime} = 2 	imes 5 = 10$. Given $SP=8$,we have $8+PS^{\prime} = 10 \implies PS^{\prime} = 2$. The distance between the foci $SS^{\prime}$ is $2ae = 2 	imes 5 	imes \frac{3}{5} = 6$. Checking the options: $4+S^{\prime}P = 4+2 = 6$. Therefore,$SS^{\prime} = 4+S^{\prime}P$.

If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and if $PSP^{\prime}$ is a focal chord with $SP=8$,then $SS^{\prime}$ is equal to

Similar Questions

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse,then the eccentricity of the ellipse is

The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

The length of the latus-rectum of the ellipse,whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$,is

Find the length of the latus rectum of the ellipse $4x^2 + 9y^2 - 36y + 4 = 0$.

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module