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 coordinates 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 coordinate 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 coordinates 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, coordinates 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.
● The Ellipse
11 and 12 Grade Math
From Problems on Ellipse to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.