Ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis

Question

Question:

Find the values of the constants $a$,$b$ and $c$ such that the ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis

What I've done so far:

Substitute $(-1,2)$ in $4x^2+y^2+ax+by+c=0$:

$$\begin{align} &\implies 4+4-a+2b+c=0 \\ &\implies c=a-2b-8 \tag{1} \end{align}$$

Rewriting the equation of the ellipse to get standard form:

$$\begin{align} & 4x^2+y^2+ax+by=-c \\[6pt] \implies& 4\left(x^2+\frac{a}{4}x+\frac{a^2}{64}\right) +\left(y^2+by+\frac{b^2}{4}\right)=a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} \\[6pt] \implies& \frac{4\left(x+\frac{a}{8}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{\left(y+\frac{b}{2}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}=1 \end{align}$$

Since the ellipse is tangent to the $x$-axis:

$$\begin{align} &\quad\sqrt{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}} = \frac{b}{2} \\[6pt] \implies&\quad a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} = \frac{b^2}{4} \\ \implies&\quad 2b= a-\frac{a^2}{16}-8 \tag{2} \end{align}$$

From $(1)$ and $(2)$:

$$\quad c= a-\left(a-\frac{a^2}{16}-8\right)- 8 \quad\implies\quad c= \frac{a^2}{16} \tag{3}$$

Therefore, the equation of the ellipse becomes

$$\frac{4\left(x+\frac{a}{8}\right)^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{(y+\frac{b}{2})^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}=1$$

That is, $$\frac{\left(x+\frac{a}{8}\right)^2}{b^2/16}+\frac{\left(y+\frac{b}{2}\right)^2}{b^2/4}=1 \tag{4}$$

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I'm stuck at this point and don't know how to find $a$ (or what else to do in general).

All help is appreciated.

Answer 1

There's not one, but an infinite number of ellipses that satisfy the two conditions that you have listed. You have 3 unknowns ($a$,$b$ and $c$) to solve for, but only 2 conditions given by which you can determine them. So, no unique solution.

Note that the equation of the ellipse can be written as $$\frac{(x+\frac{a}{8})^2}{(\frac{1}{2})^2} + \frac{(y+\frac{b}{2})^2}{1^2} = \frac{a^2}{4^2}+\frac{b^2}{2^2}+c \equiv D^2$$.

Some edits: I forgot to divide out by $D^2$ to put it in standard form. So, $$\frac{(x+\frac{a}{8})^2}{(\frac{D}{2})^2} + \frac{(y+\frac{b}{2})^2}{D^2} = 1$$. This shows that the half axis along the y-direction has length D. Since the ellipse is tangent to the x-axis, the center of the ellipse $(\frac{-a}{8},\frac{-b}{2}$ must be exactly D above the x-axis. In other words, $\frac{-b}{2}=D$ or $\frac{b^2}{4}=D^2$.
$$\frac{a^2}{4^2}+\frac{b^2}{2^2}+c = \frac{b^2}{4}$$ This is one equation relating $a$,$b$ and $c$. The other is obtained by plugging $(x,y)=(-1,2)$ in to the equation of the ellipse to get the second equation

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

Any combination of $a$, $b$ and $c$ that satisfy the two equations above, will give a valid ellipse that meets the two conditions you are given.

Answer 2

Your equation $(1)$ is correct: $$a - 2 b - c = 8 \tag{1}$$

For the tangency condition, you're thinking a little too hard. If the ellipse is tangent to the $x$-axis, then the equation should have a double-root for $x$ when $y=0$; that is, its discriminant must vanish: $$4 x^2 + a x + c = 0 \quad\to\quad \text{discriminant} = a^2 - 4\cdot 4\cdot c = 0 \quad\to\quad a^2 = 16 c \tag{2}$$

... which is, of course, your $(3)$. Now, two independent equations aren't enough to specify three unknowns. There are lots of solutions here. Since $a^2 \geq 0$, we know $c$ is non-negative; take $c := (2d)^2$ for some $d$ (the factor of $2$ helps us avoid some fractions), and we have

$$a = \pm 8 d \qquad b = -2( d^2 \mp 2 d + 2 ) \qquad c = 4d^2 \tag{3}$$

For completeness, the equation of the ellipse is ... $$4 \left(x \pm d \right)^2 + \left( y -(d^2\mp 2 d+2) \right)^2 = (d^2\mp 2 d + 2)^2 \tag{4}$$

Note that the vertical radius is $\left|d^2\mp 2 d + 2\right|$, which matches the vertical offset of the center, easily re-confirming the tangency property. $\square$