Find the co-ordinates of the point of intersaction of T and N?

**Jaguis** · March 22nd 2009, 05:32 AM

The tangent to the curve y=x^2 + 4x - 5 at (1,0) is T and N is the normal to the curve y = x^2 - x + 2 at (1,2)

**Soroban** · March 22nd 2009, 05:49 AM

Hello, Jaguis!

It's quite straight-forward . . . Exactly where is your difficulty?

The tangent to the curve $y\:=\:x^2 + 4x - 5$ at (1,0) is $T$
and $N$ is the normal to the curve $y \:=\: x^2 - x + 2$ at (1,2)

Find the intersection of $T$ and $N$ .

The slope of the tangent to $y \:=\:x^2+4x-5$ is: . $y' \:=\:2x+4$

At (1,0), the slope is: . $m_T \:=\:6$

The equation of the tangent is: . $y - 0 \:=\:6(x-1) \quad\Rightarrow\quad y \:=\:6x-6$ .[1]

The slope of the tangent to $y \:=\:x^2-x+2$ is: . $y' \:=\:2x-1$

At (1,2), the slope is: . $m_T \:=\:1$

Hence, the slope of the normal is: . $m_N = -1$

The equation of the normal is: . $y - 2 \:=\:-1(x-1) \quad\Rightarrow\quad y \:=\:-x + 3$ .[2]

Now find the intersection of [1] and [2] . . .

**Jaguis** · March 22nd 2009, 05:56 AM

ok thanks. well calculus is not my main area of study in university but i have a study unit on elementary calculus and differentiation is just a little bit confusing.... at least for now.

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Find the co-ordinates of the point of intersaction of T and N?

thanks

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