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:
2(3)+9(2)+k=0, k=-24
2(-3)+9(-2)+k=0, k=24
So k=+24,-24
Awesome, thanks for the help!
Well, determining the values of k for which the line is normal to the hyperbola requires a careful analysis. But before we get into the serious stuff, let's try to lighten the mood a little!
Why did the hyperbola go to therapy? Because it had an identity crisis!
Now, let's get back to business. To find the values of k, we need to consider the relationship between the line and the hyperbola.
The first step is to find the slope of the tangent line to the hyperbola at any point (x, y). Taking the derivative of the equation of the hyperbola, we get:
6x - 2yy' = 0
Simplifying, we find:
y' = 3x/y
Now that we've got the slope, we can determine the slope of the line perpendicular to the tangent line by taking the negative reciprocal:
m = -y/x
Since the line 2x + 9y + k = 0 is normal to the hyperbola, its slope must be the negative reciprocal of the slope we found earlier.
Now, let's try to interject some humor while we continue with the calculations. Just bear with me, this might get a little tangential!
Why was the tangent line upset with the hyperbola? Because it didn't invite it to any of its parties!
Back to the problem at hand! We have the equation of the line: 2x + 9y + k = 0. As we found earlier, the slope of this line is -9/2.
So, we have the equation:
-9/2 = -y/x
To simplify, we cross-multiply:
2y + 9x = 0
Now, we express y in terms of x:
y = -9x/2
Comparing this equation with the equation of the line, we can see that they are identical when k = -9/2.
Therefore, for the line 2x + 9y + k = 0 to be normal to the hyperbola, the value of k must be -9/2.
Hope that enlightened you while solving, and remember: always maintain your sense of humor, even in math problems!
To determine the values of k for which the line is normal to the hyperbola, we need to consider the properties of normal lines to conic sections. In particular, for a line to be normal to a hyperbola at a certain point, the slope of the line must be the negative reciprocal of the slope of the tangent line at that point.
First, let's find the equation of the hyperbola. We have the equation 3x^2 - y^2 = 23. Rearranging this equation, we get y^2 = 3x^2 - 23.
To find the slope of the tangent line at a point (x0, y0) on the hyperbola, we need to take the derivative of the equation and evaluate it at that point. Differentiating both sides of the equation, we get:
2y * dy/dx = 6x
dy/dx = 3x/y
Now, let's find the slope of the tangent line at the point of intersection between the hyperbola and the line 2x + 9y + k = 0. To find this point, we substitute the equation of the line into the equation of the hyperbola:
3x^2 - (-(2x + k)/9)^2 = 23
3x^2 - (4x^2 + 4kx + k^2)/81 = 23
243x^2 - 4x^2 - 4kx - k^2 - 2079 = 0
239x^2 - 4kx - k^2 - 2079 = 0
This quadratic equation in x should have equal roots for the line to be tangent to the hyperbola. Therefore, we can set the discriminant of this equation to zero:
(-4k)^2 - 4(239)(- k^2 - 2079) = 0
16k^2 + 4(239)(k^2 + 2079) = 0
16k^2 + 956k^2 + 1985160 = 0
972k^2 + 1985160 = 0
Solving this quadratic equation, we get two values for k.