Number of solutions of a parametric equation

Question

Find the number of solutions of the equation \begin{equation*} \sqrt{1-4x^2}=mx+m+\frac{1}{2}, \end{equation*} where $m$ is a real parameter.

The solutions to this exercise should be: one solution for $-1 \le m \le -\frac{1}{3}$, two solutions for $-\frac{1}{3} < m \le \frac{-2+\sqrt{13}}{3}$, no solutions otherwise.

Please note: this exercise can be solved by studying the number of interception points between the line $y=m$ and the function $y=\frac{\sqrt{1-4x^2}-\frac{1}{2}}{1+x}$. I am not allowed to use this method. The solution has to be found using one of the following two attempts.

My attempts: first of all, we require that $-\frac{1}{2}\le x \le \frac{1}{2}$, so that the square root exists.

Algebraic attempt: Since we have a square root on the LHS, we also require that $mx+m+\frac{1}{2} \ge 0$. Whenever this condition is valid, we can square both sides and get \begin{equation*} m^2x^2+m^2+\frac{1}{4}+2m^2x+m+mx+4x^2-1=0. \end{equation*} By studying the discriminant of this equation, we should get the number of solutions it has. One gets that $\Delta=0 \iff m = \frac{-2\pm\sqrt{13}}{3}$. Isn't this telling us there only are two values of $m$ for which the parametrical equation has one solution only? Moreover: by studying $\Delta \ge 0$ , we get $m \le \frac{-2-\sqrt{13}}{3} \lor m \ge \frac{-2+\sqrt{13}}{3}$, which doesn't agree with the expected answer.

Geometrical attempt: We can find points where the curves $y=\sqrt{1-4x^2}$ and $y=mx+m+\frac{1}{2}$ intercept. The first curve is a semiellipse in the plane. Its equation is $4x^2+y^2=1$, with the restriction $y\ge 0$. The second curve is a line. Finding the interception points between the line and the ellipse should provide the solutions to the exercise. By writing a system of equations, I end up (... of course) in the exact same computations I get in the algebraic method.

Possible problems: I think the (one of the) problem(s) lies in the fact I am not really using the condition $y \ge 0$, hence I am finding the interception points between the line and the whole ellipse.

Answer 1

The algebraic approach is mechanical: $$\sqrt{1-4x^2}=mx+m+\frac12\iff$$ $$mx+m+\frac12\ge0\quad\text{and}\quad1-4x^2=\left(mx+m+\frac12\right)^2.$$ (There is no need to "keep in mind that the solution must fall inside the range $-\frac12\le x\le\frac12$": this will automatically be the case since $1-4x^2$ will be a square.) $$1-4x^2=\left(mx+m+\frac12\right)^2\iff$$ $$(m^2+4)x^2+m(2m+1)x+m^2+m-\frac34=0.$$ $$\Delta:=-4(3m^2+4m-3)\ge0\iff m\in I:=\left[\frac{-2-\sqrt{13}}3,\frac{-2+\sqrt{13}}3\right].$$ For $m\in I,$ the two potential solutions are $$x_\pm:=\frac{-m(2m+1)\pm\sqrt\Delta}{2(m^2+4)}.$$

The additional condition $mx_\pm+m+\frac12\ge0$ is satisfied iff $$0\le-m^2(2m+1)\pm m\sqrt\Delta+(2m+1)(m^2+4),$$ i.e. $$\mp m\sqrt\Delta\le4(2m+1).$$ Now, $$4^2(2m+1)^2-m^2\Delta=12(m+1)\left(m+\frac13\right)(m^2+4).$$ Therefore:

• if $m<-1$ or $m>\frac{-2+\sqrt{13}}3,$ there is no solution;
• if $-1\le m\le-\frac13$ or $m=\frac{-2+\sqrt{13}}3,$ there is one solution: $x_-;$
• if $-\frac13<m<\frac{-2+\sqrt{13}}3,$ there are two solutions: $x_\pm.$

Answer 2

The discriminant of the squared equation is $\Delta=-4(3m^2+4m-3)$, which has roots $ \frac{-2\pm\sqrt{13}}{3}$ . You seem to have an error in the denominator.

One thing that you should keep in mind is that the original equation involves a $\sqrt{1-4x^2}$ and thus is only valid for $-\frac{1}{2} \leq x \leq \frac{1}{2}$.

Therefore, when do the solution of the squared equations fall outside the range $-\frac{1}{2} \leq x \leq \frac{1}{2}$ ?

Answer 3

The geometrical approach is little easier to understand (and helps in understanding the algebraic method).

First, let us consider the full ellipse, $y^2 = 1 - 4x^2$. Now consider, the intersection between the ellipse and $mx + m + 1/2$. The domain of $m$, will resemble something similar to $t_1 \le m \le t_2$, whence $m = t_1,t_2$, there is $1$ intersection and $t_1 < m < t_2$ there is $2$ intersections.

With a little calculus, it is not too hard to find $t_1, t_2$. Since, $1$ intersection indicates tangency, the following shows up

$$\begin{cases} \sqrt{1-4x^2} = mx + m + 1/2 \\ \big(\sqrt{1-4x^2}\big)'= \dfrac{-4x}{\sqrt{1-4x^2}}=m\end{cases}$$

Rearranging both in terms of $m$:

$$\begin{cases} \dfrac{\sqrt{1-4x^2} - 1/2}{x + 1} = m \\ \big(\sqrt{1-4x^2}\big)'= \dfrac{-4x}{\sqrt{1-4x^2}}=m\end{cases}$$

Essentially, $\dfrac{\sqrt{1-4x^2} - 1/2}{x + 1} = \dfrac{-4x}{\sqrt{1-4x^2}}$, which, with some manipulation leads to $17x^2 + 8x + \frac{3}{4} = 0$

$\therefore x = \dfrac{-8 \pm \sqrt{13}}{34}$ and $t_1 = \dfrac{-2 - \sqrt{13}}{3}, t_2 = \dfrac{-2 + \sqrt{13}}{3}$

Key note: $t_1$ is the tangency for $y < 0$ (negative ellipse), and should not exist as a solution under the positive restriction.

Currently, we have analysed the "1 solution" part, but not completely as we assumed a full ellipse. Now, consider when the line intersects the full ellipse twice. There are going combinations of $-$ and $+$ solutions.

1. For some $m$, both intersections are $y > 0$, which is correctly considered 2 solutions w.r.t your question

2. Other $m$, one intersection is $y > 0$ and other is $y < 0$, which while two solutions, is only $1$ solution w.r.t your question as the semi-ellipse for $y < 0$ does not exist.

3. Lastly, there are $m$ for which both intersection are $y < 0$, hence do not count as solutions.

Given the following criteria above and the ellipse restriction between $-1/2 \le x \le 1/2$, the domain of $m$ would resemble:

$u \le m \le k \space \cup \space m = \dfrac{-2 + \sqrt{13}}{3}$ for $1$ solution and $k < m < \dfrac{-2 + \sqrt{13}}{3}$ for $2$ solution. $k$ is very special value ;), because that is when the line "leaves" the ellipse for $x > 0$, precisely at $x = \dfrac{1}{2}, y = 0$ and hence the change between 2 solutions to 1 solution.

Hence, evaluating $m_k$ at $x = \dfrac{1}{2}$:

$$\sqrt{1 - 4(1/2)^2} = m_k(1/2 + 1) + 1/2 \\ m_k = \dfrac{-1/2}{1/2 + 1} = \dfrac{-1}{3}$$

The value, $m = u$, would be when the line completely leaves the ellipse, and that is at $x = -\dfrac{1}{2}, y = 0$. (Note that the line always existed towards the left of the ellipse, and hence the extent of the rotation arc for $x > 0$ is far greater than $x < 0$.)

Hence, evaluating $m_u$ at $x = \dfrac{-1}{2}$

$$\sqrt{1 - 4(-1/2)^2} = m_u(-1/2 + 1) + 1/2 \\ m_u = \dfrac{-1/2}{-1/2 + 1} = -1$$

$$\therefore m := \begin{cases} \left[-1, \dfrac{-1}{3}\right] \space \cup \space \left\{ \dfrac{-2 + \sqrt{13}}{3} \right\} & 1 \text{ solution} \\ \left[\dfrac{-1}{3}, \dfrac{-2 + \sqrt{13}}{3}\right) & 2 \text{ solutions}\end{cases}$$