An ellipse having the coordinate axes as its axes and its major axis along $Y$-axis, passes through the…

An ellipse having the coordinate axes as its axes and its major axis along $Y$-axis, passes through the point $(-3,1)$ and has eccentricity $\sqrt{\frac{2}{5}}$. Then its equation is
  1. $3 x^2+5 y^2-15=0$
  2. $5 x^2+3 y^2-32=0$
  3. $3 x^2+5 y^2-32=0$
  4. $5 x^2+3 y^2-48=0$

Solution

Let the equation of ellipse is $ \begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & e=\sqrt{\frac{2}{5}} \\ & \Rightarrow \quad \sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{2}{5}} \\ & \Rightarrow \quad \frac{a^2-b^2}{a^2}=\frac{2}{5} \\ & \Rightarrow \quad 5 a^2-5 b^2=2 a^2 \\ & \end{aligned} $ Given, $ \begin{array}{ll} \Rightarrow & 3 a^2=5 b^2 \\ \Rightarrow & a^2=\frac{5 b^2}{3} \end{array} $ Now, ellipse passes through $(-3,1)$ $ \begin{aligned} & \Rightarrow \frac{(-3)^2}{a^2}+\frac{(1)^2}{b^2}=1 \Rightarrow 9 b^2+a^2=a^2 b^2 \\ & \Rightarrow 9 b^2+\frac{5 b^2}{3}=\frac{5 b^4}{3} \Rightarrow \frac{32 b^2}{3}=\frac{5 b^4}{3} \\ & \Rightarrow b^2=\frac{32}{5} \end{aligned} $ From Eq. (i), we get $ \begin{array}{rlrl} \Rightarrow & & \frac{x^2}{32}+\frac{y^2}{\frac{32}{3}} & =1 \\ \Rightarrow & 3 x^2+5 y^2 & =32 \\ \Rightarrow & & 3 x^2+5 y^2-32 & =0 . \end{array} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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