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Miscellaneous Exercise 7(I) · Q111

Q.Eccentricity of the hyperbola 16x2−3y2−32x−12y−44=016x^2 - 3y^2 - 32x - 12y - 44 = 0 is
A) 17/3\sqrt{17}/3 B) 19/3\sqrt{19}/3 C) 19/3\sqrt{19}/3 D) 17/3\sqrt{17}/3

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16x2−3y2−32x−12y−44=016x^2-3y^2-32x-12y-44=0.

Group: 16(x2−2x)−3(y2+4y)=4416(x^2-2x)-3(y^2+4y)=44.

Complete the square: 16(x2−2x+1−1)−3(y2+4y+4−4)=4416(x^2-2x+1-1)-3(y^2+4y+4-4)=44

⇒16(x−1)2−16−3(y+2)2+12=44\Rightarrow 16(x-1)^2-16-3(y+2)^2+12=44

⇒16(x−1)2−3(y+2)2=48\Rightarrow 16(x-1)^2-3(y+2)^2=48.

Divide by 48: (x−1)23−(y+2)216=1\dfrac{(x-1)^2}{3}-\dfrac{(y+2)^2}{16}=1. So a2=3, b2=16a^2=3,\,b^2=16. …

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