Finding intersection of tangent and line

**StaryNight** · October 27th 2008, 09:09 AM

I have to find the coordinates of the point of intersection of the tangents to the graph of y=x^2 at the points at which it meets the line with equation y=x+2.

This is what I have so far:
y=x^2 intersects with y=x+2 at -1 or 2 since x^2=x+2 factorises to (x+1)(x-2)

The derivative of y=x^2 is 2x , so the gradients of y=x^2 at these points are -2 and 4

I am stuck on where to proceed from here, could somebody please nudge me in the right direction?

Thanks

**Soroban** · October 27th 2008, 10:25 AM

Hello, StaryNight!

Find the coordinates of the point of intersection of the tangents to $y\:=\:x^2$
at the points at which it meets the line $y\:=\:x+2$

Your preliminary work is correct!

The slope of the tangent is given by: . $\frac{dy}{dx} = 2x$

The parabola and line intersect at: . $P(2,4)\,\text{ and }\,Q(\text{-}1,1)$

At $P(2,4)$ , the slope is: $m = 4$
The equation of the tangent is: . $y - 4 :=\:4(x-2)\quad\Rightarrow\quad y \:=\:4x-4$

At $Q(\text{-}1,1)$ , the slope is: $m = \text{-}2$
The equation of the tangent is: . $y - 1 \:=\:\text{-}2(x+1) \quad\Rightarrow\quad y \:=\:\text{-}2x - 1$

Now find where the two tangents intersect . . .

**StaryNight** · October 27th 2008, 10:44 AM

Many thanks Soroban, it turns out I misunderstood the question.

My solution is (0.5,-2)

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