Question:

If the co-ordinates of two points \(A\) and \(B\) are \(\left(\sqrt7,0\right)\) and \(\left(-\sqrt7,0\right)\) respectively and \(P\) is any point on the conic, \(9x^2+16y^2=144\), then \(PA+PB\) is equal to :

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Convert to \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and check that \(A,B\) are its foci; then \(PA+PB=2a\).
Updated On: Nov 5, 2026
  • \(8\)
  • \(9\)
  • \(16\)
  • \(6\)
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The Correct Option is A

Solution and Explanation

Divide by 144: \(\frac{x^2}{16}+\frac{y^2}{9}=1\), so \(a=4\), \(b=3\).
\(e^2=1-\frac{b^2}{a^2}=1-\frac9{16}=\frac7{16}\Rightarrow ae=\sqrt7\). So the foci are \((\pm\sqrt7,0)\), i.e. \(A\) and \(B\).
By the focal property of an ellipse, \(PA+PB=2a=8\).
Answer: (1) \(8\)
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