# Latus Rectum of Ellipse

It is a chord of the ellipse passing through a focus and perpendicular to the major axis.

Let LL' be the latus rectum of the ellipse

(1)

passing through the focus F

_{2 }(ae, 0).

The coordinates of L are (ae, F

_{2}L).

Since L lies on (1)

Hence, length of the lactus rectum is .

**Example**

**Find the eccentricity of the ellipse****If its latus rectum is half of its minor-axis,****If its latus rectum is half of its major axis.**

**Solution**

- Let the equation of the ellipse be
- Its latus rectum = and its minor axis = 2b

Given latus rectum = (minor axis)

or a = 2b

Thus, - Latus rectum = and major axis = 2a.

Given latus rectum = (major axis)

â‡’ 2b^{2 }= a^{2}or b^{2 }=

Thus,

- Its latus rectum = and its minor axis = 2b
**Find an equation of the ellipse referred to its axes as the coordinate axes with latus rectum of length 4 and distance between foci .**

**Solution**

Let the equation of the ellipse be

Latus rectum = = 4 or b

^{2 }= 2a -------------------(1)

Distance between the foci ---------------(2)

Since b

^{2}= a

^{2 }(1 âˆ’ e

^{2}) â‡’ 2a = a

^{2 âˆ’ }a

^{2 }e

^{2 }= a

^{2 âˆ’ }8 [using (2)]

â‡’ a

^{2 âˆ’ }2a - 8 = 0 or (a âˆ’ 4) (a + 2) = 0 â‡’ a = 4, âˆ’ 2

But a cannot be negative, therefore, a = 4 â‡’ b

^{2 }= 2a = 8

Hence, equation of the ellipse is

**Find the eccentricity of the ellipse****if the distance between the foci is equal to the length of the latus rectum.****if the minor axis is equal to the distance between the foci.**

**Solution**

- Distance between the foci = length of the latus rectum

Since for an ellipse, 0<, e<1, we reject

- Minor axis = distance between the foci â‡’ 2b = 2ae or b
^{2 }= a^{2}e^{2}

or a^{2}(1 âˆ’ e^{2}) = a^{2 }e^{2}or 1 âˆ’ e^{2 }= e^{2}or 2e^{2 }= 1 â‡’ .

**Find an equation of the ellipse whose axes are long the coordinate axes, centre at the origin, latus rectum = 8 and****.**

**Solution**

Latus rectum = = 8 â‡’ b

^{2 }= 4a.

Since , from the relation b

^{2 }= a

^{2 }(1 âˆ’ e

^{2}), we get or or a = 8.

Thus, b

^{2 }= (4)(8) = 32.

Hence, equation of the required ellipse is

**Find the equation of the ellipse whose minor axis is equal to the distance between the foci and whose latus rectum is 10.**

**Solution**

We are given 2b = 2ae or b = ae

â‡’ b

^{2 }= a

^{2 }e

^{2}â‡’ a

^{2 }(1 âˆ’ e

^{2}) = a

^{2 }e

^{2}

â‡’ 1 âˆ’ e

^{2 }= e

^{2}or 2e

^{2 }= 1 or .

Also, since the latus rectum = = 10, we get b

^{2 }= 5a â‡’ a

^{2 }(1 âˆ’ e

^{2}) = 5a

â‡’ or a = 10. Thus, b

^{2 }= 5 (10) = 50.

Hence, equation of the required ellipse is

**Find an equation of the set of all points whose distance from (0,4) is of their distance from the line y = 9.**

**â€‹**

**Solution**

Let P(x, y) be any point on the ellipse. By definition distance between P and (0, 4) is

**times of its distance from the line y = 9.**

â‡’ 9[x

^{2 }+ y

^{2 âˆ’ }8y + 16] = 4(y

^{2 âˆ’ }18y + 81)

â‡’ 9x

^{2 }+ 5y

^{2 }= 180.