# Thread:

1. ## The equation of the normal to a curve which is parallel to a given line

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

2. Originally Posted by mastermin346
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

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.

3. ## WRITE it

Originally Posted by mr fantastic
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.

can u write it?it dificult to me.

4. Originally Posted by mastermin346
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

i am sure you know that the gradient of $y=1/2x+3/2$ is 1/2

$y=-x^2+2x+3$

$\frac{dy}{dx}=-2x+2$ ... gradient of tangent

so gradient of normal will be $-\frac{1}{2-2x}$

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 .

5. Originally Posted by mastermin346
can u write it?it dificult to me.
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?
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