<s>[INST] 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 $(-1,2)$ 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?
 [/INST] 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 = $\frac{\sqrt3}{2}$
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