TS EAMCET · Maths · Ellipse
If \(S\) and \(S^{\prime}\) are the foci of an ellipse \(\frac{x^2}{169}+\frac{y^2}{144}=1\) and the point \(B\) lying on positive Y-axis is one end of its minor axis, then the incentre of the triangle SBS' is
- A \(\left(0, \frac{10}{3}\right)\)
- B \(\left(\frac{13}{3}, \frac{10}{3}\right)\)
- C \(\left(\frac{10}{3}, \frac{13}{3}\right)\)
- D \(\left(0, \frac{13}{3}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(0, \frac{10}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(a^2=169, b^2=144 \Rightarrow a=13, b=12\) \(c^2=a^2-b^2=169-144=25 \Rightarrow c=5\) \(S=(5,0), S'=(-5,0), B=(0,12)\) Side lengths of \(\triangle SBS'\): \(s_{BS} = \sqrt{(5-0)^2+(0-12)^2}=13\) \(s_{BS'} = \sqrt{(-5-0)^2+(0-12)^2}=13\)…
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