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)
Hello, Jaguis!
It's quite straight-forward . . . Exactly where is your difficulty?
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] . . .