AP EAMCET · Maths · Ellipse
The center of ellipse \(x^2+2 y^2-4 x+12 y+14=0\) is
- A \((-2,-3)\)
- B \((-2,3)\)
- C \((2,-3)\)
- D \((2,6)\)
Answer & Solution
Correct Answer
(C) \((2,-3)\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & & x^2+2 y^2-4 x+12 y+14=0 \\ \Rightarrow & & (x-2)^2+2(y+3)^2=-14+4+18 \\ \Rightarrow & & (x-2)^2+2(y+3)^2=8 \\ \Rightarrow & & \frac{(x-2)^2}{8}+\frac{(y+3)^2}{8 / 2}=1 \end{aligned} \] \(\therefore \quad\) Centre of ellipse is \((2,-3)\).
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