If the focal distance of an endpoint of the m | vedclass

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

Question

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The distance between the foci is $2ae = 2h$,so $ae = h$.T

Vedclass · Accepted Answer

$\frac{x^2}{k^2} + \frac{y^2}{k^2 - h^2} = 1$

Vedclass · Answer

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.The distance between the foci is $2ae = 2h$,so $ae = h$.The endpoints of the minor axis are $(0, b)$ and $(0, -b)$.The focal distance of an endpoint of the minor axis is the distance from $(0, b)$ to a focus $(ae, 0)$,which is $\sqrt{(ae - 0)^2 + (0 - b)^2} = \sqrt{a^2e^2 + b^2}$.Given this distance is $k$,we have $\sqrt{a^2e^2 + b^2} = k$,so $a^2e^2 + b^2 = k^2$.Substituting $ae = h$,we get $h^2 + b^2 = k^2$,which implies $b^2 = k^2 - h^2$.We also know $b^2 = a^2(1 - e^2) = a^2 - a^2e^2 = a^2 - h^2$.Thus,$a^2 - h^2 = k^2 - h^2$,which gives $a^2 = k^2$.Substituting $a^2 = k^2$ and $b^2 = k^2 - h^2$ into the standard equation,we get $\frac{x^2}{k^2} + \frac{y^2}{k^2 - h^2} = 1$.

If the focal distance of an endpoint of the minor axis of an ellipse (taking its axes as the $x$ and $y$ axes respectively) is $k$ and the distance between its foci is $2h$,then its equation is:

Similar Questions

In an ellipse,let $B$ be one end of the minor axis,$F$ and $F'$ be the foci,and $\angle FBF' = 90^{\circ}$. Then the eccentricity of the ellipse is:

$A$ line passing through the endpoint of the major axis $A$ and the endpoint of the minor axis $B$ of an ellipse $\frac{x^2}{9} + y^2 = 1$ touches its auxiliary circle at point $M$. Find the area of the triangle with vertices $A, M,$ and the origin $O$.

The coordinates of the foci of the ellipse $16x^{2} + 9y^{2} = 144$ are

Consider an ellipse with foci at $(5, 15)$ and $(21, 15)$. If the $X$-axis is a tangent to the ellipse,then the length of its major axis equals

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