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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to :
Solution
For hyperbola, latus rectum length $=\dfrac{2b^2}{a}=8\Rightarrow b^2=4a$. Since $P(10,2\sqrt{15})$ lies on it, $\dfrac{100}{a^2}-\dfrac{60}{b^2}=1$. Using $b^2=4a$, we get $a=5$, so $b^2=20$ and $c^2=a^2+b^2=45$. Hence $SS'=2c=6\sqrt5$. Area of $\Delta PSS' = \dfrac12\cdot 6\sqrt5\cdot 2\sqrt{15}=30\sqrt3$. Therefore square of area $=(30\sqrt3)^2=2700$.Online Test Series, Information About Examination,
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