Name: Latus Rectum of an Ellipse
Uploaded: 2019-07-19
Duration: 2 min 46 s
Description: Latus rectum is a special chord of a conic. An Ellipse has two Latus recta. To learn how to draw the latus rectum and know its length, watch the video!

Question

Show that the area of sector of the ellipse between the major axis and a line drawn through the focus is equal to

\frac{1}{2} a b (θ - e \sin θ)

sq units, where

θ

is the eccentric angle of the point to which the line is drawn through the focus and

e

is the eccentricity of the ellipse.

Accepted Answer

To prove that the area of sector of the ellipse of semi axes $a$ and $b$ between the major axis and a radius vector from the focus, where $θ$ is the eccentric angle of the point to which the radius vector is drawn.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ . $O$ is the centre and $S (a e, 0)$ is the focus. Let $θ$ be the eccentric angle of any point $P (x, y)$ on the ellipse.

Then $x = a \cos θ, y = b \sin θ$ .

$S P$ is the radius vector of $P$ drawn through the focus $S$ and $S A$ is the radius vector along the major axis.At the point $A, x = a$ and $θ = 0$ .

Then we draw $P M$ perpendicular to the $x$ -axis.

Then the required area of the sector $S A P =$ Area of the traingle $S M P +$ area $P M A$

$= \frac{1}{2} \cdot S M \cdot M P + \int_{a \cos θ}^{a} y d x$ , for the ellipse

$= \frac{1}{2} (O M - O S) \cdot M P + \int_{0}^{θ} y \cdot \frac{d x}{d θ} \cdot d θ$

$= \frac{1}{2} (a \cos θ - a θ) \cdot b \sin θ + \int_{0}^{θ} b \sin θ \cdot (- a \sin θ) d θ$

Since $x = a \cos θ & y = b \sin θ$

$= \frac{1}{2} a b (\cos θ - e) \sin θ + \int_{0}^{θ} a b \cdot \frac{1}{2} (1 - \cos 2 θ) d θ$
$= \frac{1}{2} a b (\cos θ - e) \sin θ + \frac{1}{2} a b {[θ - \frac{\sin 2 θ}{2}]}_{0}^{θ}$

$= \frac{1}{2} a b (\cos θ - e) \sin θ + \frac{1}{2} a b (θ - \frac{1}{2} \cdot 2 \cdot \sin θ \cdot \cos θ)$

$= \frac{1}{2} a b [\cos θ \cdot \sin θ - e \cdot \sin θ + θ - \sin θ \cdot \cos θ]$

$= \frac{1}{2} a b (θ - e \sin θ)$

Show that the area of sector of the ellipse between the major axis and a line drawn through the focus is equal to $\frac{1}{2} a b (θ - e \sin θ)$ sq units, where $θ$ is the eccentric angle of the point to which the line is drawn through the focus and $e$ is the eccentricity of the ellipse.

Important Questions on Ellipse

Learn and Prepare for any exam you want !

Show that the area of sector of the ellipse between the major axis and a line drawn through the focus is equal to 12abθ-esinθ sq units, where θ is the eccentric angle of the point to which the line is drawn through the focus and e is the eccentricity of the ellipse.

Important Questions on Ellipse

Learn and Prepare for any exam you want !

Show that the area of sector of the ellipse between the major axis and a line drawn through the focus is equal to $\frac{1}{2} a b (θ - e \sin θ)$ sq units, where $θ$ is the eccentric angle of the point to which the line is drawn through the focus and $e$ is the eccentricity of the ellipse.