Foci of the ellipse \(2 x^2+3 y^2-4 x-12 y+13=0\) are
Foci of the ellipse \(2 x^2+3 y^2-4 x-12 y+13=0\) are
- \(\left(1+\frac{1}{\sqrt{6}}, 2\right)\) and \(\left(1-\frac{1}{\sqrt{6}}, 2\right)\)
- \(\left(\frac{1}{\sqrt{6}}+1,2\right)\) and \(\left(\frac{1}{\sqrt{6}}-1,2\right)\)
- \(\left(2,1+\frac{1}{\sqrt{6}}\right)\) and \(\left(2,1-\frac{1}{\sqrt{6}}\right)\)
- \(\left(2, \frac{1}{\sqrt{6}}+1\right)\) and \(\left(2, \frac{1}{\sqrt{6}}-1\right)\)
Solution
Given ellipse
\(\begin{array}{rlrl}
& & 2 x^2+3 y^2-4 x-12 y+13 & =0 \\
\Rightarrow & 2(x-1)^2+3(y-2)^2 & =1 \\
\Rightarrow & & \frac{(x-1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2}+\frac{(y-2)^2}{\left(\frac{1}{\sqrt{3}}\right)^2} & =1
\end{array}\)
\(\begin{aligned}
& e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1 / 3}{1 / 2}}=\frac{1}{\sqrt{3}} \\
& a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{3}} \Rightarrow a e=\frac{1}{\sqrt{6}}
\end{aligned}\)
Foci \(\quad(h \pm a e, k)=\left(1 \pm \frac{1}{\sqrt{6}}, 2\right)\)
Asked in: AP EAMCET 2020 (17 Sep Shift 1)
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