Question

The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes through the points (-3, 1) and (2, -2) is

• A. $5x^2+3y^2=32$
• B. $3x^2+5y^2=32$
• C. $5x^2-3y^2=32$
• D. $3x^2+5y^2+32=0$

Answer 1

The correct option is B $3x^2+5y^2=32$

Standard equation of ellipse : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

it passes through (-3,1) and (2,-2)

when passing through (-3,1)

$\frac{{(-3)}^2}{a^2}+\frac{{\left(1\right)}^2}{b^2}=1$

${9b}^2+a^2=a^2{b}^2⟶\left(1\right)$

when passing through (2,-2)

$\frac{{\left(2\right)}^2}{a^2}+\frac{{(-2)}^2}{b^2}=1$

${4a}^2+{4b}^2=a^2{b}^2⟶\left(2\right)$

$eqn\left(1\right)=eqn\left(2\right)\Rightarrow {9b}^2+a^2={4a}^2+{4b}^2$

$3a^2={5b}^2$

$b^2=\frac{3}{5}{a}^2⟶\left(3\right)$

Now, eqn becomes,

$\frac{x^2}{a^2}+\frac{5y^2}{{3a}^2}=1$

$\frac{{3x}^2+5y^2}{{3a}^2}=1$

${3x}^2+{5y}^2={3a}^2⟶\left(4\right)$

when passing through (-3,1)

$3\times {(-3)}^2+5\times 1=3a^2$

$27+5={3a}^2$

${3a}^2=32$$⟶\left(5\right)$

putting value of $3a^2$ in equation(4)

$3x^2+{5y}^2=32$ Answer