Question

The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which asses through the points (–3, 1) and (2, –2) is
(a) 5x² + 3y² = 32
(b) 3x² + 5y² = 32
(c) 5x² – 3y² = 32
(d) 3x² + 5y² + 32 = 0

Answer 1

for an ellipse, centre is given as (0, 0) major axis as x axis
i.e equation is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{where}a>b$
Since ellipses passes through (–3, 1) and (2, –2)
We get,
$$\begin{gathered} \frac{9}{a^2}+\frac{1}{b^2}=1\mathrm{and}\frac{4}{a^2}+\frac{4}{b^2}=1 \\ \mathrm{i}.\mathrm{e}-\frac{9}{a^2}+1=\frac{1}{b^2}\mathrm{and}\frac{4}{a^2}+4\left(\frac{1}{b^2}\right)=1 \\ \mathrm{i}.\mathrm{e}\frac{4}{a^2}+4\left(1-\frac{9}{a^2}\right)=1 \\ \mathrm{i}.\mathrm{e}\frac{4}{a^2}+4-\frac{36}{a^2}=1 \\ \mathrm{i}.\mathrm{e}\frac{4-36}{a^2}=1-4 \\ \mathrm{i}.\mathrm{e}\frac{-32}{a^2}=-3 \\ \mathrm{i}.\mathrm{e}\frac{1}{a^2}=\frac{3}{32} \end{gathered}$$
$$\begin{gathered} \therefore \frac{1}{b^2}=1-\frac{9}{a^2} \\ =1-\frac{9\times 3}{32} \\ \frac{1}{b^2}=\frac{32-27}{32} \\ \mathrm{i}.\mathrm{e}\frac{1}{b^2}=\frac{5}{32} \\ \mathrm{Therefore}\mathrm{equation}\mathrm{of}\mathrm{ellipse}\mathrm{is}\frac{3x^2}{32}+\frac{5y^2}{32}=1 \\ \mathrm{i}.\mathrm{e}3x^2+5y^2=32 \end{gathered}$$

Hence, the correct answer is option C.