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Q.Write the equation of the directrices of the hyperbola x216−y29=1\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Subjective· 1mImportance★★★★★
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With a=4a=4, b=3b=3, c=5c=5, the eccentricity is e=5/4e=5/4, giving directrices x=±16/5x=\pm16/5.

Comparing x216−y29=1\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 with the standard form x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1: a2=16⇒a=4a^2=16 \Rightarrow a=4, and b2=9⇒b=3b^2=9 \Rightarrow b=3.

For a hyperbola, c2=a2+b2=16+9=25⇒c=5c^2 = a^2+b^2 = 16+9=25 \Rightarrow c=5.

Eccentricity e=ca=54e = \dfrac{c}{a} = \dfrac{5}{4}.

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