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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $S$ and $S'$ be the foci of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$ and $P(\alpha,\beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S'P)^2-SP\cdot S'P=37$, then $\alpha^2+\beta^2$ is equal to :
Solution
For the ellipse, $a=5$, $b=3$, so $c=4$. Let $SP=r_1$, $S'P=r_2$. Since $P$ lies on ellipse, $r_1+r_2=2a=10$. Given $r_1^2+r_2^2-r_1r_2=37$. Using $(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2=100$, we get $3r_1r_2=63\Rightarrow r_1r_2=21$, and hence $r_1^2+r_2^2=58$. But $r_1^2+r_2^2=2(\alpha^2+\beta^2+c^2)=2(\alpha^2+\beta^2+16)$. Therefore $2(\alpha^2+\beta^2+16)=58\Rightarrow \alpha^2+\beta^2=13$.Online Test Series, Information About Examination,
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