Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (- 3, 1) and has eccentricity
.
Given that we need to find the equation of the ellipse (whose axes are x = 0 and y = 0) which passes through the point (- 3,1) and has eccentricity
.

We know that the equation of the ellipse whose axes are x and y - axis is
. ..... - - - - - (1)
Let us assume a2>b2.
We know that eccentricity(e) = ![]()
⇒ ![]()
⇒ ![]()
⇒ ![]()
⇒
..... .... (2)
Substituting (2) in (1) we get,
⇒ ![]()
⇒ ![]()
⇒ 3x2 + 5y2 = 3a2
This curve passes through the point (- 3,1). Substituting in the curve we get,
⇒ 3(- 3)2 + 5(1)2 = 3a2
⇒ 3(9) + 5 = 3a2
⇒ 32 = 3a2
⇒ ![]()
⇒ ![]()
⇒ ![]()
The equation of the ellipse is:
⇒ ![]()
⇒ ![]()
⇒ 3x2 + 5y2 = 32
∴ The equation of the ellipse is 3x2 + 5y2 = 32.
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