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Exercise 7.3 · Q74

Q.Find the equation of the hyperbola with centre at the origin, length of conjugate axis 10 and one of the foci (−7,0)(-7,0).

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2b=10⇒b=5⇒b2=252b=10 \Rightarrow b=5 \Rightarrow b^2=25.

Focus (−7,0)(-7,0) gives ae=7⇒a2e2=49ae=7 \Rightarrow a^2e^2=49.

Since a2e2=a2+b2a^2e^2=a^2+b^2: 49=a2+25⇒a2=2449=a^2+25 \Rightarrow a^2=24. …

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