Question:
On the ellipse \(9x^2 + 25y^2 = 225\), find the point, the distance from which to the focus \(F_2\) is four times the distance to the focus \(F_1\).
On the ellipse \(9x^2 + 25y^2 = 225\), find the point, the distance from which to the focus \(F_2\) is four times the distance to the focus \(F_1\).
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Packing fraction measures the deviation of nuclear mass from integer atomic mass.
Updated On: June 02, 2025
- \(\left(-\frac{15}{4}, \frac{\sqrt{63}}{4} \right)\)
- \(\left(\frac{15}{4}, \frac{\sqrt{63}}{4} \right)\)
- \(\left(-\frac{1}{15}, \frac{-\sqrt{63}}{4} \right) \text{ and } \left(-\frac{1}{15}, \frac{\sqrt{63}}{4} \right)\)
- \(\left(\frac{1}{15}, \frac{-\sqrt{63}}{4} \right) \text{ and } \left(\frac{1}{15}, \frac{\sqrt{63}}{4} \right)\)
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The Correct Option is B
Solution and Explanation
Ellipse: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\) ⇒ semi-major axis \(a = 5\), semi-minor \(b = 3\) Foci: \((\pm c, 0)\), where \[ c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \] Let point be \((x, y)\). Distance to \(F_1 = \sqrt{(x + 4)^2 + y^2}\) Distance to \(F_2 = \sqrt{(x - 4)^2 + y^2}\) Given: \[ \sqrt{(x - 4)^2 + y^2} = 4 \sqrt{(x + 4)^2 + y^2} \] Solve to get \((x, y) = \left(\frac{15}{4}, \frac{\sqrt{63}}{4} \right)\)
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