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