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  • The equation <img alt="\mathrm{16 x^2-3 y^2-32 x-12 y-44=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B16%20x%5E2-3%20y%5E2-32%20x-12%20y-44%3D0%7D

The equation \mathrm{16 x^2-3 y^2-32 x-12 y-44=0} represents a hyperbola 

 

Option: 1

the length of whose transverse axis is 2 \sqrt{3}


Option: 2

the length of whose conjugate axis is 8

 


Option: 3

whose centre is at (1, −2)


Option: 4

all of these 

 


Answers (1)

\mathrm{\text { Since } 16 x^2-32 x=16(x-1)^2-16 \text { and } 3 y^2+12 y=3(y+2)^2-12}

 the given equation can be written \mathrm{16(x-1)^2-16-3(y+2)^2+12-44=0}

\mathrm{\begin{aligned} & \Rightarrow \quad 16(x-1)^2-3(y+2)^2-48=0 \\ & \quad \frac{(x-1)^2}{3}-\frac{(y+2)^2}{16}=1 \end{aligned}}

Comparing it to the standard equation of a hyperbola centered at the origin,\mathrm{\frac{x^2}{a^2}=\frac{y^2}{b^2}=1}  , we see that the given curve is a hyperbola with centre at (1, −2), and for which \mathrm{a=\sqrt{3}} and b = 4. Thus the length of the transverse axis is \mathrm{2 a=2 \sqrt{3}} ; the length of the conjugate axis is 2b = 8 ; and the eccentricity is

\mathrm{e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{3}}=\sqrt{\frac{19}{3}}}

 

 

Posted by

Sumit Saini

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