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.

● 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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