KCET · Maths · Ellipse
If \(P\) is any point on the ellipse \(\frac{x^{2}}{36}+\frac{y^{2}}{16}=1\) and \(S\) and \(S^{\prime}\) are the foci, then \(P S+P S^{\prime}\) is equal to
- A 4
- B 8
- C 10
- D 12
Answer & Solution
Correct Answer
(D) 12
Step-by-step Solution
Detailed explanation
Given ellipse is \(\frac{x^{2}}{36}+\frac{y^{2}}{16}=1\)
Here, \(\quad a^{2}=36, b^{2}=16\)
Since, \(a>b\), so the sum of the focal distance of any point \(P\) on the ellipse is \(P S=P S^{\prime}+2 a\)
\(\Rightarrow \quad P S+P S^{\prime}=2 \times 6\)
\(\Rightarrow \quad P S+P S^{\prime}=12\)
Here, \(\quad a^{2}=36, b^{2}=16\)
Since, \(a>b\), so the sum of the focal distance of any point \(P\) on the ellipse is \(P S=P S^{\prime}+2 a\)
\(\Rightarrow \quad P S+P S^{\prime}=2 \times 6\)
\(\Rightarrow \quad P S+P S^{\prime}=12\)
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