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)

- Mar 22nd 2009, 05:32 AMJaguisFind the co-ordinates of the point of intersaction of T and N?
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)

- Mar 22nd 2009, 05:49 AMSoroban
Hello, Jaguis!

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

Quote:

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

and $\displaystyle N$ is the normal to the curve $\displaystyle y \:=\: x^2 - x + 2$ at (1,2)

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

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

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

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

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

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

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

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

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

- Mar 22nd 2009, 05:56 AMJaguisthanks
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.