Q53 / 125 · BITSAT 2012 · Mathematics

Find the eccentricity of the conic represented by $x^{2}-y^{2}-4 x+4 y+16=0$

  • A
    2
  • Correct answer
    B
    $\sqrt{2}$
  • C
    $2 \sqrt{2}$
  • D
    $3 \sqrt{2}$

Solution

We have $x^{2}-y^{2}-4 x+4 y+16=0$ $\Rightarrow\left(x^{2}-4 x\right)-\left(y^{2}-4 y\right)=16$ $\Rightarrow\left(x^{2}-4 x+4\right)-\left(y^{2}-4 y+4\right)=-16$ $\Rightarrow(x-2)^{2}-(y-2)^{2}=-16$ $\Rightarrow \frac{(x-2)^{2}}{4^{2}}-\frac{(y-2)^{2}}{4^{2}}=1$ This is rectangular hyperbola, whose eccentricity is always $\sqrt{2}$.

Hence, the correct answer is option: 'B'.

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