Question:
Let \(O(0, 0)\) and \(A(0, 1)\) be two fixed points. Then the locus of a point P such that the perimeter of \(\Delta AOP\) is 4, is :
Let \(O(0, 0)\) and \(A(0, 1)\) be two fixed points. Then the locus of a point P such that the perimeter of \(\Delta AOP\) is 4, is :
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Since \(OA = 1\), the condition says \(PO + PA = 3\), a constant: the locus is an ellipse with foci O and A.
Updated On: Nov 5, 2026
- \(9x^2 + 8y^2 - 8y = 16\)
- \(8x^2 - 9y^2 + 9y = 18\)
- \(8x^2 + 9y^2 - 9y = 18\)
- \(9x^2 - 8y^2 + 8y = 16\)
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The Correct Option is A
Solution and Explanation
\(OA = 1\), perimeter \(= 4\) \(\Rightarrow\) \(PO + PA = 3\), an ellipse with foci \(O(0, 0)\), \(A(0, 1)\).
Centre \(= \left(0, \frac{1}{2}\right)\), \(2a = 3 \Rightarrow a = \frac{3}{2}\), \(2c = 1 \Rightarrow c = \frac{1}{2}\), \(b^2 = a^2 - c^2 = \frac{9}{4} - \frac{1}{4} = 2\).
Major axis is along the y-axis: \(\frac{x^2}{2} + \frac{(y - 1/2)^2}{9/4} = 1\).
\(\Rightarrow 9x^2 + 8\left(y^2 - y + \frac{1}{4}\right) = 18 \Rightarrow 9x^2 + 8y^2 - 8y = 16\).
Answer: (1) \(9x^2 + 8y^2 - 8y = 16\)
Centre \(= \left(0, \frac{1}{2}\right)\), \(2a = 3 \Rightarrow a = \frac{3}{2}\), \(2c = 1 \Rightarrow c = \frac{1}{2}\), \(b^2 = a^2 - c^2 = \frac{9}{4} - \frac{1}{4} = 2\).
Major axis is along the y-axis: \(\frac{x^2}{2} + \frac{(y - 1/2)^2}{9/4} = 1\).
\(\Rightarrow 9x^2 + 8\left(y^2 - y + \frac{1}{4}\right) = 18 \Rightarrow 9x^2 + 8y^2 - 8y = 16\).
Answer: (1) \(9x^2 + 8y^2 - 8y = 16\)
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