Question

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

• A.
• B.
• C.
• D.

Answer 1

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

Let the required ellipse equation be $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$.....(i)
Where, $(h,k)$ is centre and $a$ and $b$ length of the semi major axis.
Given centre is $(h,k)=(0,0)$
Substitute $(h,k)=(0,0)$ in equation (i), we get
$\Rightarrow \frac{(x-0{)}^2}{a^2}+\frac{(y-0{)}^2}{b^2}=1$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .....(ii)
Since it passes through (-3,1) and (2,-2) ,
So $\frac{9}{a^2}+\frac{1}{b^2}=1$ .....(iii) and
$\frac{4}{a^2}+\frac{4}{b^2}=1$
$\Rightarrow \frac{1}{a^2}=\frac{1}{4}-\frac{1}{b^2}$
Substitute above equation in equation(iii), we get
$\Rightarrow \frac{9}{4}-\frac{9}{b^2}+\frac{1}{b^2}=1$
$\Rightarrow -\frac{8}{b^2}=1-\frac{9}{4}$
$=\frac{4-9}{4}$
$\Rightarrow -\frac{8}{b^2}=\frac{-5}{4}$
$\Rightarrow \frac{1}{b^2}=\times \frac{5}{4\times 8}$
$\therefore b^2=\frac{5}{32}$
Substitute $b^2$ value in $\frac{1}{a^2}=\frac{1}{4}-\frac{1}{b^2}$ to get value of $a^2$.
$\Rightarrow \frac{1}{a^2}=\frac{1}{4}-\frac{1}{\frac{32}{5}}$
$\Rightarrow \frac{1}{a^2}=\frac{1}{4}-\frac{5}{32}$
$=\frac{8-5}{32}$
$\therefore \frac{1}{a^2}=\frac{3}{32}$
Substitute the values of $\frac{1}{a^2}=\frac{3}{32}$ and $b^2=\frac{5}{32}$ in equation (ii), we get
$\frac{3}{32}\times x^2+\frac{5}{32}\times y^2=1$

Hence, required equation of ellipse is $3x^2+5y^2=32$

Trick : Since only equation $3x^2+5y^2=32$ passes through (-3,1) and (2,-2). Hence the result .