JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
If the co-ordinates of two points \(A\) and \(B\) are \((\sqrt{7}, 0)\) and \((-\sqrt{7}, 0)\) respectively and \(P\) is any point on the conic, \(9 x^{2}+16 y^{2}=144,\) then \(PA + PB\) is equal to
- A \(8\)
- B \(6\)
- C \(16\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) \(a =4 ; b =3 ; e =\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\) A and \(B\) are foci \(\Rightarrow PA + PB =2 a =2 \times 4=8\)
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