Q97 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Mathematics
Find the equation of the ellipse which passes through the points $(-3,1)$ and $(2,-2)$, whose center lies at $(0,0)$ and major axis lies along the $X$-axis.
- Correct answerA$3 x^2+5 y^2=32$
- B$5 x^2+3 y^2=32$
- C$5 x^2-3 y^2=32$
- D$3 x^2+5 y^2=132$
Solution
$\therefore$ Centre and major axis of an ellipse are $(0,0)$ and $X$-axis respectively.
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be the equation of ellipse.
According to question it will pass through $(-3,1)$ and $(2,-2)$
$$
\Rightarrow \quad \frac{9}{a^2}+\frac{1}{b^2}=1...(i)
$$
and $\frac{4}{a^2}+\frac{4}{b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{4}...(ii)$
Subtract Eq. (ii) from Eq. (i), we get
$$
\frac{9}{a^2}-\frac{1}{a^2}=1-\frac{1}{4}
$$
$$
\Rightarrow \quad \frac{8}{a^2}=\frac{3}{4} \Rightarrow a^2=\frac{32}{3}
$$
By Eq. (i), we get
$$
\begin{array}{rlrl}
& & \frac{9}{\frac{32}{3}}+\frac{1}{b^2} & =1 \\
\Rightarrow & & \frac{27}{32}+\frac{1}{b^2} & =1 \Rightarrow \frac{1}{b^2}=1-\frac{27}{32} \\
\Rightarrow & \frac{1}{b^2} & =\frac{32-27}{32} \\
\Rightarrow & \frac{1}{b^2}=\frac{5}{32} \Rightarrow b^2 & =\frac{32}{5}
\end{array}
$$
$\begin{array}{ll}\therefore \text { Equation of ellipse } \frac{x^2}{\frac{32}{3}}+\frac{y^2}{\frac{32}{5}}=1 \\ \Rightarrow & \frac{3 x^2}{32}+\frac{5 y^2}{32}=1 \\ \Rightarrow & 3 x^2+5 y^2=32\end{array}$
Hence, the correct answer is option: 'A'.