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Latus Rectum - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Length of Latusrectum is considered one of the most asked concept.

  • 24 Questions around this concept.

Solve by difficulty

Consider an ellipse, whose centre is at the origin and its major axis is along the x-axis.  If its eccentricity is \small \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 the latus rectum of the ellipse \mathrm{x^2 \tan ^2 \alpha+y^2 \sec ^2 \alpha=1} is \frac{1}{2}, then \alpha(0<\alpha<\frac{\pi }{3}) is equal to:

In an ellipse the distance between its foci is 6 and its minor axis is 8. Then its eccentricity is 

The distance of the point \mathrm{' \theta^{\prime}}  on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} from a focus is

Concepts Covered - 1

Length of Latusrectum

Length of Latusrectum

\\ {\text {Let Latus rectum } \mathrm{LL}^{\prime}=2 \alpha} \\ {\mathrm{S}(\mathrm{ae}, 0) \text { is focus, then } \mathrm{LS}=\mathrm{SL}^{\prime}=\alpha} \\ {\text {Coordinates of } \mathrm{L} \text { and } \mathrm{L}^{\prime} \text { become }(\mathrm{ae}, \alpha) \text { and }(\mathrm{ae},-\alpha) \text { respectively }} \\ {\text {Equation of ellipse, } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1} \\ {Put\,\,x=ae,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)} \\ {\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 Points of Latus rectum

\mathrm{L}=\left(\mathrm{ae}, \frac{\mathrm{b}^{2}}{\mathrm{a}}\right) \text { and } \mathrm{L}^{\prime}=\left(\mathrm{ae},-\frac{\mathrm{b}^{2}}{\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 P(x, y) be any point on the ellipse.

 

\\ {\text { Here, \;\;\;\;\;\;\;\;\;} \quad S P=e P M=e\left(\frac{a}{e}-x\right)=a-e x} \\\\ {\quad \quad\quad\quad\quad\quad\; S^{\prime} P=e P M'=e\left(\frac{a}{e}+x\right)=a+e x}

Now, SP + S’P = a – ex + a + ex = 2a = AA' =  constant. 

Thus the sum of the focal distances of a point on the ellipse is constant.

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Length of Latusrectum

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