this is the question:
find the equation of the normal to the curve y=3+2x-x^2 which is parallel to the line 2y-x-3=0
please help me...i dont know how to do..
The gradient of 2y-x-3=0 is 1/2. Therefore the gradient of the required normal is 1/2 therefore the gradient of the tangent is -2. So solve dy/dx = -2 to get the x-coordinate and hence y-coordinate of the required point on the curve. Now you have a point and you have a gradient and therefore you can write down the equation of the normal line.
i am sure you know that the gradient of $\displaystyle y=1/2x+3/2$ is 1/2
$\displaystyle y=-x^2+2x+3$
$\displaystyle \frac{dy}{dx}=-2x+2$ ... gradient of tangent
so gradient of normal will be $\displaystyle -\frac{1}{2-2x}$
have it equal to 1/2 . Do you know why ?
then solve for x , sub into the original equation to get y.
After that , make use the formula that i gave you in my post tp your other question to get the equation .
You have spent no more than 8 minutes thinking about what I posted before aksing for more help. You need to spend more time than that thinking about the question and the reply I gave.
Have you tried doing what I suggested? What don't you understand? Where do you get stuck?