"Find the equation of the ellipse x2+9y2−4x−72y+139=0 and..." was answered by our community

balff1t

Answered question

2021-11-14

Find the equation of the ellipse

x^{2} + 9 y^{2} - 4 x - 72 y + 139 = 0

and locate the coordinates of the foci.

Answer & Explanation

Nancy Johnson

Beginner2021-11-15Added 17 answers

Ellipse is basically a plane curve around 2 focal points such that for all points on the curve, the sum of the two distances to the focal points is a constant. General equation of ellipse is given by
$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$
A circle is a special type of ellipse in which both the focal points are same.
Step 2
Given equation of ellipse: $x^{2} + 9 y^{2} - 4 x - 72 y + 139 = 0$
Completing the squares of x and y terms:
$x^{2} - 4 x + 9 y^{2} - 72 y = - 139$
$\Rightarrow x^{2} - 4 x + 4 + 9 y^{2} - 72 y = - 139 + 4$
$\Rightarrow (x^{2} - 4 x + 4) + 9 (y^{2} - 8 y + 16) = - 139 + 4 + 144$
$\Rightarrow {(x - 2)}^{2} + 9 {(y - 4)}^{2} = 9$
$\Rightarrow \frac{{(x - 2)}^{2}}{9} + {(y - 4)}^{2} = 1$
Step 3
Center of the ellipse is $(2, 4) = (h, k)$
Eccentricity of the ellipse will be:
$\frac{\sqrt{a^{2} - b^{2}}}{a} = \frac{\sqrt{3^{2} - 1^{2}}}{3} = \frac{\sqrt{9 - 1}}{3} = \frac{\sqrt{8}}{3} = \frac{2 \sqrt{2}}{3}$
Now, focii of ellipse: $(h - c, k), (h + c, k)$
where,
$c = \sqrt{a^{2} - b^{2}} = \sqrt{9 - 1} = \sqrt{8} = 2 \sqrt{2}$
Hence, Focii will be:
$(2 - 2 \sqrt{2}, 4)$ and $(2 + 2 \sqrt{2}, 4)$

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