Latus Rectum of the Ellipse

We will discuss about the latus rectum of the ellipse along with the examples.

Definition of the latus rectum of an ellipse:

The chord of the ellipse through its one focus and perpendicular to the major axis (or parallel to the directrix) is called the latus rectum of the ellipse.

It is a double ordinate passing through the focus. Suppose the equation of the ellipse be $\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1 then, from the above figure we observe that L$_{1}$SL$_{2}$ is the latus rectum and L$_{1}$S is called the semi-latus rectum. Again we see that M$_{1}$SM$_{2}$ is also another latus rectum.

According to the diagram, the co-ordinates of the end L$_{1}$ of the latus rectum L$_{1}$SL$_{2}$ are (ae, SL$_{1}$). As L$_{1}$ lies on the ellipse $\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1, therefore, we get,

$\frac{(ae)^{2}}{a^{2}}$ + $\frac{(SL_{1})^{2}}{b^{2}}$ = 1

$\frac{a^{2}e^{2}}{a^{2}}$ + $\frac{(SL_{1})^{2}}{b^{2}}$ = 1     

e$^{2}$ + $\frac{(SL_{1})^{2}}{b^{2}}$ = 1

⇒ $\frac{(SL_{1})^{2}}{b^{2}}$ = 1 - e$^{2}$

⇒ SL$_{1}$$^{2}$ = b$^{2}$ . $\frac{b^{2}}{a^{2}}$, [Since, we know that, b$^{2}$ = a$^{2}$(1 - e$^{2}$)]

⇒ SL$_{1}$$^{2}$ = $\frac{b^{4}}{a^{2}}$       

Hence, SL$_{1}$ = ± $\frac{b^{2}}{a}$.

Therefore, the co-ordinates of the ends L$_{1}$ and L$_{2}$ are (ae, $\frac{b^{2}}{a}$) and (ae, - $\frac{b^{2}}{a}$) respectively and the length of latus rectum = L$_{1}$SL$_{2}$ = 2 . SL$_{1}$ = 2 . $\frac{b^{2}}{a}$ = 2a(1 - e$^{2}$)

Notes:

(i) The equations of the latera recta of the ellipse $\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1 are x = ± ae.

(ii) An ellipse has two latus rectum.

Solved examples to find the length of the latus rectum of an ellipse:

Find the length of the latus rectum and equation of the latus rectum of the ellipse x$^{2}$ + 4y$^{2}$ + 2x + 16y + 13 = 0.

Solution:

The given equation of the ellipse x$^{2}$ + 4y$^{2}$ + 2x + 16y + 13 = 0

Now form the above equation we get,

(x$^{2}$ + 2x + 1) + 4(y$^{2}$ + 4y + 4) = 4

⇒ (x + 1)$^{2}$ + 4(y + 2)$^{2}$ = 4.

Now dividing both sides by 4

⇒ $\frac{(x + 1)^{2}}{4}$ + (y + 2)$^{2}$ = 1.

⇒ $\frac{(x + 1)^{2}}{2^2} + \frac{(y + 2)^{2}}{1^{2}}$ ………………. (i)

Shifting the origin at (-1, -2) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have

x = X - 1 and y = Y - 2 ………………. (ii)

Using these relations, equation (i) reduces to $\frac{X^{2}}{2^{2}}$ + $\frac{Y^{2}}{1^{2}}$ = 1 ………………. (iii)

This is of the form $\frac{X^{2}}{a^{2}}$ + $\frac{Y^{2}}{b^{2}}$ = 1, where a = 2 and b = 1.

Thus, the given equation represents an ellipse.

Clearly, a > b. So, the given equation represents an ellipse whose major and minor axes are along X and Y axes respectively.

Now fine the eccentricity of the ellipse:

We know that e = $\sqrt{1 - \frac{b^{2}}{a^{2}}}$ = $\sqrt{1 - \frac{1^{2}}{2^{2}}}$ = $\sqrt{1 - \frac{1}{4}}$ = $\frac{√3}{2}$.

Therefore, the length of the latus rectum = $\frac{2b^{2}}{a}$ = $\frac{2 ∙ (1)^{2}}{2}$ = $\frac{2}{2}$ = 1.

The equations of the latus recta with respect to the new axes are X= ±ae

X = ± 2 ∙ $\frac{√3}{2}$

⇒ X = ± √3

Hence, the equations of the latus recta with respect to the old axes are

x = ±√3 – 1, [Putting X = ± √3 in (ii)]

i.e., x = √3 - 1 and x = -√3 – 1.

● The Ellipse

• Definition of Ellipse
• Standard Equation of an Ellipse
• Two Foci and Two Directrices of the Ellipse
• Vertex of the Ellipse
• Centre of the Ellipse
• Major and Minor Axes of the Ellipse
• Latus Rectum of the Ellipse
• Position of a Point with respect to the Ellipse
• Ellipse Formulae
• Focal Distance of a Point on the Ellipse
• Problems on Ellipse

11 and 12 Grade Math 

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