Question

Find the ellipse satisfying the following conditions:

1. Vertex (±5, 0); foci (±4, 0)

2. Ends of the major axis (±3, 0), ends of minor axis (0, ±2)

3. Length of the major axis 26, foci (±5, 0)

4. b = 3, c = 4, centre at origin; foci on the x-axis

Answer 1

1. Foci (±4, 0) lie on the x-axis. So the equation of the ellipse is of the form \(\frac{x^2}{a} + \frac{y^2}{b^2} = 1\)

Given; Vertex (±5, 0) ⇒ a = 5

Given; Foci(±4, 0) Foci ⇒ c = 4 = \(\sqrt{a^2 - b^2}\)

⇒ 4 = \(\sqrt{25-b^2}\) ⇒ 16 = 25 – b² ⇒ b² = 9

Therefore the equation of the ellipse is

\(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

2. The ends of major axis lie on the x-axis. So the equation of the ellipse is of the form 

 \(\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1\)

Given; Ends of the major

axis (±3, 0) ⇒ a = 3, ends of minor axis

(0, ±2) ⇒ b = 2

Therefore the equation of the ellipse is 

\(\frac{x^2}{9} + \frac{y^2}{4} = 1.\)

3. Since foci (±5, 0) lie on x-axis, the standard form of ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) 

Given; 2a = 26 ⇒ a = 13

Given; c = 5 = \(\sqrt{a^2-b^2}\)

⇒ 25 = 169 – b² ⇒ b² = 144

Therefore the equation of the ellipse is

\(\frac{x^2}{169} + \frac{y^2}{144} = 1.\)

4. The standard form of ellipse is

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

Given; c = 4 = \(\sqrt{a^2-b^2}\)

⇒ 16 = a² – 9 

⇒ a² =25

Therefore the equation of the ellipse is

\(\frac{x^2}{25} + \frac{y^2}{9} = 1\).