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  • The eccentricity of the conic represented by  <img alt="\mathrm{x^2-y^2-4 x+4 y+16=0 \text { is }}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2-y%5E2-4%20x

The eccentricity of the conic represented by 

\mathrm{x^2-y^2-4 x+4 y+16=0 \text { is }}

Option: 1

1


Option: 2

\mathrm{\sqrt{2}}


Option: 3

2


Option: 4

1/2


Answers (1)

best_answer

We have, \mathrm{x^2-y^2-4 x+4 y+16=0}

\mathrm{\begin{aligned} & \Rightarrow\left(x^2-4 x\right)-\left(y^2-4 y\right)=-16 \\\\ & \Rightarrow\left(x^2-4 x+4\right)-\left(y^2-4 y+4\right)=-16 \\\\ & \Rightarrow(x-2)^2-(y-2)^2=-16 \\\\ & \Rightarrow \frac{(x-2)^2}{4^2}-\frac{(y-2)^2}{4^2}=-1 \end{aligned}}

Shifting the origin at (2,2), we obtain 

\mathrm{\frac{X^2}{4^2}-\frac{Y^2}{4^2}=-1 \text {, where } x=X+2, y=Y+2 \text {. }}

Clearly, it is a rectangular hyperbola, whose eccentricity

is always \mathrm{\sqrt{2} .}

Posted by

Riya

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