MHB Problem of the Week #112 - May 19th, 2014

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: A lemniscate curve is defined implicitly by the equation $(x^2+y^2)^2=4(x^2-y^2)$.

1. Find the coordinates of the four points at which the tangent line is horizontal.
2. At what points is the tangent line vertical?
3. Find the equation of the two tangent lines to the curve at $x=0$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by lfdahl and MarkFL. You can find lfdahl's solution below.

[sp]I have tried to solve the POTW #112 without using polar coordinates.
\[(1). \;\;\;\;(x^2+y^2)^2=4(x^2-y^2)\]
(1) Implicit differentiation yields:

\[2(x^2+y^2)(2x+2yy')=4(2x-2yy')\\\\ \Rightarrow x(x^2+y^2)+yy'(x^2+y^2)=2x-2yy'\\\\ \Rightarrow (2).\;\;\;\;y'=\frac{x(2-x^2-y^2)}{y(2+x^2+y^2)}\]

Horizontal tangents:

\[y' = 0 \Rightarrow x^2+y^2=2\]

Using $(1)$:

\[(x^2+y^2)^2=2^2 = 4(x^2-y^2)\Rightarrow x^2-y^2 =1\Rightarrow 2x^2 = 3\Rightarrow x= \pm\sqrt{\frac{3}{2}}\\\\ y^2 = 2-x^2 = \frac{1}{2}\Rightarrow y = \pm \frac{1}{\sqrt{2}}\]

Horizontal tangents in:

\[\left ( \sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}} \right ), \left ( \sqrt{\frac{3}{2}},-\frac{1}{\sqrt{2}} \right ), \left ( -\sqrt{\frac{3}{2}},\frac{1}{\sqrt{2}} \right ), \left ( -\sqrt{\frac{3}{2}},-\frac{1}{\sqrt{2}} \right )\]

(2) Vertical tangents: $\frac{\mathrm{dx} }{\mathrm{d} y}=0$. Implicit differentiation (and letting x’=0) yields:

\[2(x^2+y^2)(2xx'+2y)=4(2xx'-2y)\Rightarrow (x^2+y^2)y=-2y\Rightarrow y = 0\]

Using $(1)$:

\[x^4 = 4x^2\Rightarrow x = \pm 2\]

Vertical tangents in:

\[\left ( -2,0 \right ), \left ( 2,0 \right )\]

(3) $x=0$ implies: $y^4=-4y^2 \Rightarrow y = 0$. Therefore, using L´Hospitals rule in $(2)$

\[y' = \frac{(2-x^2-y^2)-2x(x+yy')}{y'(2+x^2+y^2)+2y(x+yy')}=\frac{1}{y'}\Rightarrow y' = \pm 1\]

Thus, the equations of the two tangent lines in $x=0$ are:

$y=x$ and $y = -x$[/sp]