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Q.Find the foci, eccentricity of the hyperbola 16y2−9x2=14416y^2 - 9x^2 = 144.

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 2mImportance★★★★★
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y29−x216=1\dfrac{y^2}{9}-\dfrac{x^2}{16}=1 gives a2=9,b2=16a^2=9,b^2=16, e=1+169=53e=\sqrt{1+\tfrac{16}{9}}=\tfrac53, foci (0,±5)(0,\pm5).

Divide 16y2−9x2=14416y^2-9x^2=144 by 144144:

y29−x216=1.\dfrac{y^2}{9}-\dfrac{x^2}{16}=1.

This is a hyperbola with transverse axis along the yy-axis, a2=9a^2=9 (the y2y^2 term), b2=16b^2=16.

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