# Draw a tangent that is perpendicular to the line

1. Sep 30, 2005

### TSN79

I have a curve $-\sqrt {2x^3 }$ on which I'm supposed to draw a tangent that is perpendicular to the line $\[y = \frac{4}{3}x + \frac{1}{3}$.

I know that this tangent must have a "steepness" of -3/4 in order to make it perpendicular, but how do I now find the point on the graph?

2. Sep 30, 2005

### EnumaElish

Solve for the x from the curve formula that would make its slope equal to that of the tangent.

3. Sep 30, 2005

### Diane_

Given that the tangent needs to be perpendicular to the line y = 4/3x + 1/3, he's right that the slope must be -3/4. Negative reciprocal for the normal line.

TSN79 - What you need is the point on the curve where the first derivative is -3/4. Just take the derivative, plug in that value, and solve for x.

Note: I'm presuming you're taking or have taken calculus, as I don't see any other way to do this.

4. Sep 30, 2005

### EnumaElish

He could do it geometrically I suppose, but that may or may not be exact.

5. Oct 1, 2005

### TSN79

Take the derivative of what? And plug in what value? -3/4? I tried with $- \sqrt {2x^3 }$ but didn't really get anywhere...

6. Oct 1, 2005

### EnumaElish

Take the derivative of $- \sqrt {2x^3 }$ then equate it to -3/4 and solve for x.

7. Oct 1, 2005

### TSN79

Ah, now where getting somewhere, thanks ya all!

8. Oct 1, 2005

### HallsofIvy

Staff Emeritus
Did you really need to be told that? You were told that the line had to be tangent to the curve given by $y=-\sqrt {2x^3 }$. Didn't you connect "tangent" with "derivative"- after you had been told the find the derivative- the only tangent mentioned was to $y= -\sqrt {2x^3 }$.
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