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Exercise 7.3 · Q80

Q.Find the equation of the hyperbola referred to its principal axes whose length of conjugate axis = 12 and passing through (11,−2)(11, -2).

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2b=12⇒b=6⇒b2=362b=12 \Rightarrow b=6 \Rightarrow b^2=36.

The hyperbola is x2a2−y236=1\dfrac{x^2}{a^2}-\dfrac{y^2}{36}=1. It passes through (11,−2)(11,-2):

121a2−436=1⇒121a2=1+19=109⇒a2=121×910=108910.\dfrac{121}{a^2}-\dfrac{4}{36}=1 \Rightarrow \dfrac{121}{a^2}=1+\dfrac19=\dfrac{10}{9} \Rightarrow a^2=121\times\dfrac{9}{10}=\dfrac{1089}{10}. …

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