Question

If x and y are real numbers such that ${{x}^{2}}+2xy-{{y}^{2}}=6$, find the minimum value of ${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}$.

Answer 1

Convert the given equation ${{x}^{2}}+2xy-{{y}^{2}}=6$ into its parametric form by assuming, $x=r\cos \theta$ and $y=r\sin \theta$. Substitute these values in the given equation and find the value of r. Now find the value of r in terms of x and y and take power four on both sides of the expression to get the answer.

Complete step by step answer:
We have been provided with the equation: - ${{x}^{2}}+2xy-{{y}^{2}}=6$. Changing this equation into its parametric form by assuming, $x=r\cos \theta$ and $y=r\sin \theta$. Substituting these values of x and y in the above expression, we get,

& \Rightarrow {{\left( r\cos \theta \right)}^{2}}+2r\cos \theta \sin \theta -{{\left( r\sin \theta \right)}^{2}}=6 \\\ & \Rightarrow {{r}^{2}}{{\cos }^{2}}\theta +2{{r}^{2}}\cos \theta \sin \theta -{{r}^{2}}{{\sin }^{2}}\theta =6 \\\ & \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)+{{r}^{2}}\left( 2\cos \theta .\sin \theta \right)=6 \\\ \end{aligned}$$ Using the identities: - $$\Rightarrow {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $$ and $$2\cos \theta \sin \theta =\sin 2\theta $$, we get, $$\begin{aligned} & \Rightarrow {{r}^{2}}\cos 2\theta +{{r}^{2}}\sin 2\theta =6 \\\ & \Rightarrow {{r}^{2}}\left( \cos 2\theta +\sin 2\theta \right)=6 \\\ & \Rightarrow {{r}^{2}}=\left( \dfrac{6}{\cos 2\theta +\sin 2\theta } \right) \\\ \end{aligned}$$ Clearly, we can see that minimum of $${{r}^{2}}$$ occurs when we have: - $$\cos 2\theta +\sin 2\theta $$ = maximum value. We know that the maximum sum of sine and cosine function is $$\sqrt{2}$$. Therefore, $$\Rightarrow {{\left( \cos 2\theta +\sin 2\theta \right)}_{\max }}=\sqrt{2}$$ $$\Rightarrow {{\left( {{r}^{2}} \right)}_{\min }}=\dfrac{6}{\sqrt{2}}$$ - (i) Now, initially we assumed that, $$x=r\cos \theta $$ and $$y=r\sin \theta $$. Squaring these two expressions and adding, we get: - $$\begin{aligned} & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) \\\ \end{aligned}$$ Using the identity: - $${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$$, we get, $$\begin{aligned} & \Rightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}} \\\ & \Rightarrow {{r}^{2}}={{x}^{2}}+{{y}^{2}} \\\ \end{aligned}$$ Therefore, substituting the value of r in equation (i), we get, $$\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}_{\min }}=\dfrac{6}{\sqrt{2}}$$ Now, squaring both sides, we get, $$\begin{aligned} & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}_{\min }={{\left( \dfrac{6}{\sqrt{2}} \right)}^{2}} \\\ & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}_{\min }=\dfrac{36}{2}=18 \\\ \end{aligned}$$ **Hence, the minimum value of expression, $${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=18$$.** **Note:** One may note that we have changed the given expression, which is a function of x and y, into the parametric form because this will make our calculation easy. This question cannot be solved by simple differentiation. One important thing to remember is that the maximum value of $$\left( \cos 2\theta +\sin 2\theta \right)$$ is $$\sqrt{2}$$. This can be found by differentiation but it will be better if one remembers the result which will save our time.