# Problems on Ellipse

We will learn how to solve different types of problems on ellipse.

1. Find the equation of the ellipse whose eccentricity is $$\frac{4}{5}$$ and axes are along the coordinate axes and with foci at (0, ± 4).

Solution:

Let the equitation of the ellipse is $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1 ……………… (i)

According to the problem, the coordinates of the foci are (0, ± 4). Therefore, we see that the major axes of the ellipse is along y axes and the minor axes of the ellipse is along x axes.

We know that the co-ordinates of the foci are (0, ±be).

Therefore, be = 4

b($$\frac{4}{5}$$) = 4, [Putting the value of e = $$\frac{4}{5}$$]

⇒ b = 5

⇒ b$$^{2}$$ = 25

Now, a$$^{2}$$ = b$$^{2}$$(1 - e$$^{2}$$)

⇒ a$$^{2}$$ = 5$$^{2}$$(1 - ($$\frac{4}{5}$$)$$^{2}$$)

⇒ a$$^{2}$$  = 25(1 - $$\frac{16}{25}$$)

⇒ a$$^{2}$$ = 9

Now putting the value of a$$^{2}$$ and b$$^{2}$$ in (i) we get, $$\frac{x^{2}}{9}$$ + $$\frac{y^{2}}{25}$$ = 1.

Therefore, the required equation of the ellipse is $$\frac{x^{2}}{9}$$ + $$\frac{y^{2}}{25}$$ = 1.

2. Determine the equation of the ellipse whose directrices along y = ± 9 and foci at (0, ± 4). Also find the length of its latus rectum.

Solution:

Let the equation of the ellipse be $$\frac{x^{2}}{a^{2}}$$ + $$\frac{y^{2}}{b^{2}}$$ = 1, ……………………………… (i)

The co-ordinate of the foci are (0, ± 4). This means that the major axes of the ellipse is along y axes and the minor axes of the ellipse is along x axes.

We know that the co-ordinates of the foci are (0, ± be) and the equations of directrices are y = ± $$\frac{b}{e}$$

Therefore, $$\frac{b}{e}$$ = 9 …………….. (ii)

and be = 4 …………….. (iii)

Now, from (ii) and (iii) we get,

b$$^{2}$$ = 36

⇒ b = 6

Now, a$$^{2}$$ = b$$^{2}$$(1 – e$$^{2}$$)

⇒ a$$^{2}$$ = b$$^{2}$$ - b$$^{2}$$e$$^{2}$$

⇒ a$$^{2}$$ = b$$^{2}$$ - (be)$$^{2}$$

⇒ a$$^{2}$$ = 6$$^{2}$$ - 4$$^{2}$$, [Putting the value of be = 4]

⇒ a$$^{2}$$ = 36 - 16

⇒ a$$^{2}$$ = 20

Therefore, the required equation of the ellipse is $$\frac{x^{2}}{20}$$ + $$\frac{y^{2}}{36}$$ = 1.

The required length of latus rectum = 2 $$\frac{a^{2}}{b}$$ = 2 $$\frac{20}{6}$$ = $$\frac{20}{3}$$ units.

3. Find the equation of the ellipse whose equation of its directrix is 3x + 4y - 5 = 0, co-ordinates of the focus are (1, 2) and the eccentricity is ½.

Solution:

Let P (x, y) be any point on the required ellipse and PM be the perpendicular from P upon the directrix 3x + 4y - 5 = 0

Then by the definition,

$$\frac{SP}{PM}$$ = e

⇒  SP = e PM

⇒ $$\sqrt{(x - 1)^{2} + (y - 2)^{2}}$$ = ½ |$$\frac{3x + 4y - 5}{\sqrt{3^{2}} + 4^{2}}$$|

⇒ (x - 1)$$^{2}$$ + (y - 2)$$^{2}$$ = ¼ $$\frac{(3x + 4y - 5)^{2}}{25}$$, [Squaring both sides]

⇒ 100(x$$^{2}$$ + y$$^{2}$$ – 2x – 4y + 5) = 9x$$^{2}$$ + 16y$$^{2}$$ + 24xy - 30x - 40y + 25

⇒ 91x$$^{2}$$ + 84y$$^{2}$$ - 24xy - 170x - 360x + 475 = 0, which is the required equation of the ellipse.

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