Q63 / 170 · Manipal (MET) 2019 · Mathematics
The length of the axis of the conic $9 x^2+4 y^2-6 x+4 y+1=0$ are
- A$\frac{1}{2}, 9$
- B$3, \frac{2}{5}$
- Correct answerC$1, \frac{2}{3}$
- D3,2
Solution
Given that equation of conic is
$\begin{aligned}
& \quad \begin{array}{l}
9 x^2+4 y^2-6 x+4 y+1=0 \\
\Rightarrow \quad 9 x^2-6 x+4 y^2+4 y+1=0 \\
\Rightarrow \quad 9\left(x-\frac{1}{3}\right)^2+4\left(y+\frac{1}{2}\right)^2=1 \\
\text { or, } \quad \frac{\left(x-\frac{1}{3}\right)^2}{\frac{1}{9}}+\frac{\left(y+\frac{1}{2}\right)^2}{\frac{1}{4}}=1
\end{array} .
\end{aligned}$
Comparing it with standard equation of ellipse $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$, we get $a=\frac{1}{3}, b=\frac{1}{2}$ Length of major axis $=2 b=2 \times \frac{1}{2}=1$ and length of minor axis $=2 a=2 \times \frac{1}{3}=\frac{2}{3}$
Comparing it with standard equation of ellipse $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$, we get $a=\frac{1}{3}, b=\frac{1}{2}$ Length of major axis $=2 b=2 \times \frac{1}{2}=1$ and length of minor axis $=2 a=2 \times \frac{1}{3}=\frac{2}{3}$
Hence, the correct answer is option: 'C'.