# Thread: Point on a Curve parallel to a line [Something Missing ?]

1. ## Point on a Curve parallel to a line [Something Missing ?]

This is a question from a worksheet my teacher printed out of a book. It seems like something is missing because I don't know what to do with the question...

The point on the curve $y = \sqrt[2]{4x+2}$ at which the normal is parallel to the line $y=-x+8$ is

1. (A) $(\frac{1}{4}, \sqrt[2]{2})$
2. (B) $(\frac{-1}{2}, 0)$
3. (C) $(0, \sqrt[2]{2})$
4. (D) $(\frac{1}{2}, 2)$
5. (E) $(1, \sqrt[2]{6})$

2. Originally Posted by StarlitxSunshine
This is a question from a worksheet my teacher printed out of a book. It seems like something is missing because I don't know what to do with the question...

The point on the curve $y = \sqrt[2]{4x+2}$ at which the normal is parallel to the line $y=-x+8$ is

1. (A) $(\frac{1}{4}, \sqrt[2]{2})$
2. (B) $(\frac{-1}{2}, 0)$
3. (C) $(0, \sqrt[2]{2})$
4. (D) $(\frac{1}{2}, 2)$
5. (E) $(1, \sqrt[2]{6})$

Nothing's missing: the normal to a point on the graph of a function is a straightline through that point which perpendicular to the tangent to the function on that point, so you must:

2) find what the slope of a NORMAL to any point of the graph is

3) make this slope equal to the given line's slope (so that they'll be parallel)

4) solve and find out what x is, and then substitute in the function and find out what the y-coordinate is.

Tonio

3. Originally Posted by tonio
Nothing's missing: the normal to a point on the graph of a function is a straightline through that point which perpendicular to the tangent to the function on that point, so you must:

2) find what the slope of a NORMAL to any point of the graph is

3) make this slope equal to the given line's slope (so that they'll be parallel)

4) solve and find out what x is, and then substitute in the function and find out what the y-coordinate is.

1). The derivative is $\frac{2}{\sqrt[2]{4x+2}}$