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 November 16th, 2012, 04:09 PM #1 Newbie   Joined: Sep 2012 Posts: 3 Thanks: 0 Finding tangent lines on a curve through a point Hi guys can you please help me out with this question: Find tangent lines on the curve x^2+4y^2=36 which pass through the point (12,3).
 November 17th, 2012, 07:39 AM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Finding tangent lines on a curve through a point This can be done implicitly, but another way is to solve the given for y. $y=\pm\sqrt{\frac{36-x^{2}}{4}}$............[1] The line that passes through (12,3) will be tangent to the lower half of the circle, $-\sqrt{\frac{36-x^{2}}{4}}$ The derivative is $\frac{x}{2\sqrt{36-x^{2}}}$ We can use $y-y_{1}=m(x-x_{1})$ and solve for x. $-\sqrt{\frac{36-x^{2}}{4}}-3=\frac{x}{2\sqrt{36-x^{2}}}(x-12)$ Solving this for x may prove daunting, but we get $x=\frac{24}{5}$ Plugging this back results in $y=\frac{-9}{5}$ So, one line is tangent at $\left(24/5, \;\ -9/5\right)$ For the top portion, we use the positive case of [1] and get $(0,3)$ As I saod, this can be done implicitly by noting the slope is $\frac{dy}{dx}=\frac{-x}{4y}$

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