I'm currently working through Beckenbach and Bellman's book "An Introduction to Inequalities." One of the questions has me a little stumped, as I'm not really sure what they're asking for.

For fixed $a, b, m$, and $n$, solve the system of equations:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and

$mx + ny = k$

choosing $k$ so that the resulting quadratic equation has a double root. I know how to solve a system of equations, but I'm not entirely sure what they're asking for here. I'm not really sure how I'd even end up with a factorable quadratic. Maybe just add the two together and try to factor?

Any help appreciated. Preferably in hint form. I'd rather not have the solution entirely spoiled.

p.s. I know what a double root is.


2 Answers 2


I would say that by the resulting quadratic the meaning is this: when you solve (e.g) $y$ from the lower equation and plug it in the first you get a quadratic in one variable.

Now these equations represent an ellipse and a line. It's nice to think it geometrically too.

EDIT: My calculations. Hope there are no errors.

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $mx + ny = k$

Assume $n\neq 0$. Then $y = \frac{k}{n} - \frac{m}{n}x$ and

$$ \frac{x^2}{a^2}+\frac{(\frac{k}{n} - \frac{m}{n}x)^2}{b^2}=1 $$ $$ (\frac{1}{a^2}+\frac{m^2}{b^2n^2})x^2-\frac{2 k m}{b^2n^2} x +\frac{k^2}{b^2 n^2}-1= 0 $$

For this to have double root, the discriminant must be zero:

$$\frac{4 k^2 m^2}{b^4n^4} - 4(\frac{1}{a^2}+\frac{m^2}{b^2n^2})(\frac{k^2}{b^2 n^2}-1) =0 $$

$$ (\frac{4 m^2}{b^4n^4} - \frac{4(\frac{1}{a^2}+\frac{m^2}{b^2n^2})}{b^2 n^2}) k^2 + 4(\frac{1}{a^2}+\frac{m^2}{b^2n^2}) =0 $$

Solving this for $k$ gives

$$ |k| =\sqrt{\frac{4(\frac{1}{a^2}+\frac{m^2}{b^2n^2})}{\frac{4(\frac{1}{a^2}+\frac{m^2}{b^2n^2})}{b^2 n^2}-\frac{4 m^2}{b^4n^4} }} = \sqrt{\frac{(\frac{1}{a^2}+\frac{m^2}{b^2n^2})} {\frac{\frac{1}{a^2}}{b^2 n^2}+\frac{m^2}{b^4 n^4}-\frac{ m^2}{b^4n^4} }}$$

$$=\frac {\sqrt{(\frac{1}{a^2}+\frac{m^2}{b^2n^2})}}{\frac{1}{abn}} = \sqrt{a^2m^2+b^2n^2}$$

  • $\begingroup$ Ok, after reading through this, the problem makes much more sense. For some reason, I never thought to think about what the discriminant would have to be. It didn't help that the wording confused me either. $\endgroup$
    – user17137
    Feb 18, 2014 at 1:19

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $x = \verts{a}\cos\pars{\theta}$ and $y = \verts{b}\sin\pars{\theta}$ satisfy the ellipse equation. So $$ m\braces{\verts{a}\cos\pars{\theta}} + n\braces{\verts{b}\sin\pars{\theta}} = k \quad\imp\quad \sin\pars{\theta} + {m\verts{a} \over n\verts{b}}\,\cos\pars{\theta} = {k \over n\verts{b}} $$

Let's $\ds{\tan\pars{\mu} = {m\verts{a} \over n\verts{b}}}$: $$\!\!\!\!\! \sin\pars{\theta}\cos\pars{\mu} + \cos\pars{\theta}\sin\pars{\mu} = {k \over n\verts{b}}\,\cos\pars{\mu} = {k \over n\verts{b}}\,{1 \over \root{\tan^{2}\pars{\mu} + 1}} ={k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}} $$

$$ \sin\pars{\theta + \mu} = {k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}} $$

If $\ds{{\verts{k} \over \root{m^{2}a^{2} + n^{2}b^{2}}} \leq 1}$ we have solutions for $\theta\ \in\ {\mathbb R}$ $\pars{~\mbox{otherwise,}\ \theta\ \in\ {\mathbb C}~}$: $$ \theta = -\arctan\pars{m\verts{a} \over n\verts{b}} + \arcsin\pars{{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}} + 2\pi n\,,\quad n\ \in\ {\mathbb Z} $$ where $0 \leq \arcsin\pars{\alpha} < 2\pi$

Once we know $\theta$ we can evaluate $x = \verts{a}\cos\pars{\theta}$ and $y = \verts{b}\sin\pars{\theta}$: \begin{align} x&=\verts{a}\cos\pars{\arctan\pars{m\verts{a} \over n\verts{b}}} \cos\pars{\arcsin\pars{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}} \\[3mm]&+ \verts{a}\sin\pars{\arctan\pars{m\verts{a} \over n\verts{b}}} \sin\pars{\arcsin\pars{k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}}} \\[3mm]&= \verts{a}\,{1 \over \root{\pars{ma/nb}^{2} + 1}}\, \root{1 - {k^{2} \over m^{2}a^{2} + n^{2}b^{2}}} \\[3mm]&+ \verts{a}\,{m\verts{a}/\pars{n\verts{b}} \over \root{\pars{ma/nb}^{2} + 1}}\, {k\sgn\pars{n} \over \root{m^{2}a^{2} + n^{2}b^{2}}} \end{align}

$$ \color{#00f}{\large% \begin{array}{rcl} x & = & {\verts{nba}\sqrt{m^{2}a^{2} + n^{2}b^{2} - k^{2}} + ma^{2}k \over m^{2}a^{2} + n^{2}b^{2}} \\[3mm] y & = & {k \over n} - {m \over n}\,x \\[3mm]&& \color{#000}{\large\mbox{when}\ {\verts{k} \over \root{m^{2}a^{2} + n^{2}b^{2}}} \leq 1} \end{array}} $$


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