Tangents intersect off of curve y = x^2 + x

theschaef

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Oct 17, 2006
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Here is the question in the text:

Find equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2+x.

I've tried a bunch of things but have got no where.. any help would be appreciated!
 
Suppose that (a,a2+a)\displaystyle \left( {a,a^2 + a} \right) is the point tangency on the curve.

Then the slope is \(\displaystyle \begin{array}{rcl}
\frac{{a^2 + a + 3}}{{a - 2}} & = & 2a + 1 \\
a^2 + a + 3 & = & 2a^2 - 3a - 2 \\
\end{array}.\)

Now solve for a,
 
Find the derivative of y=x2+x\displaystyle y=x^{2}+x

y=2x+1\displaystyle y'=2x+1. There's your slope m.

Use yy1=m(xx1)\displaystyle y-y_{1}=m(x-x_{1})

x1=2\displaystyle x_{1}=2 and y1=3\displaystyle y_{1}=-3

x2+x+3=(2x+1)(x2)\displaystyle x^{2}+x+3=(2x+1)(x-2)

Solve for x.

You can then use those values and your given coordinates to find your lines. Take a stab at it and write back if you get stuck.
 
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