TS EAMCET · Maths · Ellipse
The centre of the ellipse \(\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1\), is
- A \((-1,2)\)
- B \((1,-2)\)
- C \((-1,-2)\)
- D \((1,2)\)
Answer & Solution
Correct Answer
(D) \((1,2)\)
Step-by-step Solution
Detailed explanation
We have, \[ \frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1 \] To determine the centre of ellipse, put \(x+y-3=0\) \[ \Rightarrow \quad x+y=3 \] and \(\quad x-y+1=0\) \[ \Rightarrow \quad x-y=-1 \] On adding Eqs. (i) and (ii), we get \[ x=1 \text { and } y=2 \] Hence, the centre of…
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