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July 29, 2014

July 29, 2014

Posted by **Anonymous** on Wednesday, October 7, 2009 at 3:42pm.

- math -
**Reiny**, Wednesday, October 7, 2009 at 5:04pmI will give you the steps.

1. find dy/dx for the hyperbola

(you should get dy/dx = 3x/y)

2. if 2x + 9y = k is a normal, then its slope is -2/9

so the slope of the tangent at the point of contact should be 9/2

3. set dy/dx = 9/2 and solve for x

4 sub in that x into the original hyperbola and solve

( I got y = ±2)

5. find the x's for each of those two y's.

6. You now have the points of contact of the tangents,**and the points of contact of the normals.**

7 sub in the points from #5 into the original normal equation to find k

There are of course two of those.

- math -
**Benevolent Ben**, Wednesday, October 7, 2009 at 5:30pmAlright, this one involves a fair bit of complicated calculus, so I'm assuming you have the basics down:

First, some definitions. A normal line is a line that is perpendicular to another function at a given point. That means that the slope of the normal line is the negative reciprocal of the derivate at a given point.

We can find the slope of the given normal line (2x+9y+k=0): slope of the normal is -2/9

This means that the derivative of 3x^(2)-y^(2)=23 at some point is equal to the negative reciprocal of (-2/9), or (9/2). Let us first find the general derivative of 3x^(2)-y^(2)=23. By doing implicit differentiation, we get dy/dx=3x/y .

So we know that 3x/y=9/2. If we do some cross multiplication, it follows that y=(2/3)x. Now we can actually substitute for why into our polynomial and solve for x. By substitute I mean do this: 3x^(2)-((2/3)x)^(2)=23 [substituting (2/3)x in for y]. When we solve this for x, we get: x=+3, -3. Because y=(2/3)x, y=+2,-2.

So now we have the two points where the derivative of our polynomial has a slope of (9/2): points (3, 2) and (-3,-2)

Now that we have these points, plug the values back into 2x+9y+k=0:

2(3)+9(2)+k=0, k=-24

2(-3)+9(-2)+k=0, k=24

So k=+24,-24

- math -
**Anonymous**, Wednesday, October 7, 2009 at 7:20pmAwesome, thanks for the help!

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