Find the eccentricity of the conic $4x^2+y^2+ax+by+c=0$, if it tangent to the $x$ axis at the origin and passes through $(-1,2)$

Question

Solving this would require three equations

(1) Tangent to x axis at origin

Substituting zeroes in all $x$ and $y$ gives $c=0$

(2) Passes through (-1,2) $$4(1)+4-a+2b=0$$ $$-a+2b=-8$$

How do I find the third equation?

Answer 1

As you mentioned, one of the points on the ellipse is (0,0). Equation of the ellipse is

$\frac{(x+\frac{a}{8})^2}{1^2}+\frac{(y+\frac{b}{2})^2}{2^2} = {(\frac{a}{8})}^2 + {(\frac{b}{4})}^2$

For ellipse equation $\frac{(x \pm h)^2}{A^2}+\frac{(y \pm k)^2}{B^2} = 1$, eccentricity of ellipse

= $\frac{\sqrt{B^2-A^2}}{B}$ (where $B \ge A$)

= $\frac{\sqrt3}{2}$ (where $A = t, B = 2t$)

Here, $t = \sqrt{{(\frac{a}{8})}^2 + {(\frac{b}{4})}^2}$

Alternatively:

Based on the fact that it is tangent to x-axis at origin, the center of the ellipse will have to be on y-axis (as it is not a slanting ellipse). That means $a = 0, c = 0, b = -4$. So equation of ellipse is $x^2 + \frac{{(y-2)}^2} {4} = 1$ with major axis (along $y$ axis) $= 4$, minor axis $= 2$.

So, eccentricity = $\frac{\sqrt3}{2}$