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.





11 and 12 Grade Math 

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