posted by Anonymous .
For what values of k will the line 2x+9y+k=0 be normal to the hyperbola given by 3x^(2)-y^(2)=23? Justify your answer
I 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.
Alright, 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:
Awesome, thanks for the help!