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Length of Latusrectum is considered one of the most asked concept.
38 Questions around this concept.
Consider an ellipse, whose centre is at the origin and its major axis is along the $x$-axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is 6 , then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the end points of major and minor axes of ellipse, is
If O is the origin and OB is the semi-minor axis of an ellipse, F1 and F2 are its foci and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is
Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S^{\prime} B S$ is a right angled triangle with right angle at B and area $\left(\Delta S^{\prime} B S\right)=8 S q$. units, then the length of a latus rectum of the ellipse is :
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If the latus rectum of the ellipse $\mathrm{x}^2 \tan ^2 \alpha+\mathrm{y}^2 \sec ^2 \alpha=1$ is $\frac{1}{2}$, then $\alpha\left(0<\alpha<\frac{\pi}{3}\right)$ is equal to:
In an ellipse the distance between its foci is and its minor axis is
. Then its eccentricity is
The distance of the point on the ellipse
from a focus is
$
\text { What is the length of the latus rectum of the ellipse: } 9 x^2+25 y^2=225
$
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The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
If the latus rectum of an ellipse is equal to half the minor axis then its eccentricity is equal to
Length of Latus rectum
Let Latus rectum $\mathrm{LL}^{\prime}=2 \alpha$
$\mathrm{S}(\mathrm{ae}, 0)$ is focus, then $\mathrm{LS}=\mathrm{SL}^{\prime}=\alpha$
Coordinates of L and $\mathrm{L}^{\prime}$ become ( $\mathrm{ae}, \alpha$ ) and (ae, $-\alpha$ ) respectively
Equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Put $x=a e, y=\alpha$ we get, $\frac{(\mathrm{ae})^2}{\mathrm{a}^2}+\frac{\alpha^2}{\mathrm{~b}^2}=1 \Rightarrow \alpha^2=\mathrm{b}^2\left(1-\mathrm{e}^2\right)$
$
\begin{aligned}
& \alpha^2=\mathrm{b}^2\left(\frac{\mathrm{~b}^2}{\mathrm{a}^2}\right) \quad\left[\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\right] \\
& \alpha=\frac{\mathrm{b}^2}{\mathrm{a}} \\
& \Rightarrow 2 \alpha=\mathrm{LL}^{\prime}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}
\end{aligned}
$
End Points of Latus rectum
$\mathrm{L}=\left(\mathrm{ae}, \frac{\mathrm{v}}{\mathrm{a}}\right)$ and $\mathrm{L}^{\prime}=\left(\mathrm{ae},-\frac{\mathrm{v}}{\mathrm{a}}\right)$
Focal Distance of a Point:
The sum of the focal distance of any point on the ellipse is equal to the major axis.
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be any point on the ellipse.
Here, $\quad \begin{aligned} & S P=e P M=e\left(\frac{a}{e}-x\right)=a-e x \\ & \\ & S^{\prime} P=e P M^{\prime}=e\left(\frac{a}{e}+x\right)=a+e x\end{aligned}$
Now, $\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}=\mathrm{a}-\mathrm{ex}+\mathrm{a}+\mathrm{ex}=2 \mathrm{a}=\mathrm{AA}^{\prime}=$ constant.
Thus the sum of the focal distances of a point on the ellipse is constant.
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