Chapter A5.2, Problem 25E

To determine

To find:the eccentricity of the ellipse.

Answer to Problem 25E

The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$

Explanation of Solution

Given:

  $4x^2+y^2+ax+by+c=0$

Calculation:

As the ellipse passes through origin,that is $\left(0,0\right)$ the value of constant in the equation should be equal to Zero.

Therefore,

  $c=0$

The ellipse passes through $\left(-1,2\right)$ , therefore $\left(-1,2\right)$ should satisfy the equation of ellipse.

This implies that $-a+2b=-8$

The ellipse is tangent to x-axis, therefore its center lies on y-axis.

This implies that $a=0$ and $b=-4$

Therefore the equation can be simplified to,

  $4x^2+y^2-4y+4=4$

It could further be simplified to,

  $x^2+{\frac{\left(y-2\right)}{4}}^2=1$

Now according to the standard symbols, $a=2$ and $b=1$

Now

  $\begin{array}{l}{c}^2=a^2-b^2\\ \text{ =4}-1\\ \text{ =3}\end{array}$

This implies that $\text{c=}\sqrt{3}$ and

  $e=\frac{c}{a}$

  $=\frac{\sqrt{3}}{2}$

Conclusion:

Therefore,the eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$