00:01
We are given an equation and we are asked to rotate the axes to eliminate the xy term in this equation.
00:11
Then we're asked to write the equation in standard form, and then we're asked to sketch the graph of the resulting equation showing both sets of axes.
00:20
The equation is xy plus 1 equals 0.
00:31
So first let's rotate the axes.
00:38
This will make the substitution x equals 0.
00:47
X prime cosine theta minus y prime sine theta and y prime equals sorry y equals x prime sine theta plus y prime pugging these in, we get x prime cosine theta minus y prime sine theta times x prime sine theta plus y prime cosine theta plus 1 equals 0.
01:28
This simplifies to x prime squared times sine theta times cosine theta minus y prime squared times sine theta times cosine theta and then we have plus x prime y prime times cosine squared theta minus x prime y prime times sine squared theta all of this plus one is equal to zero so now we want to find theta so that we can eliminate the x prime y prime term.
02:37
To define the correct value of theta, we want to rotate the axes through an angle theta, where the cotangent of 2 theta is equal to a, the coefficient of x squared.
02:58
This was 0, minus c, the coefficient of y squared.
03:04
This was also 0 over b, the coefficient of of x y, which was 1.
03:12
So the cotangent of 2 theta is simply equal to 0.
03:16
This implies that 2 theta must be equal to pi over 2, so that theta is equal to pi over 4, or 45 degrees.
03:40
Now we can complete our substitution so that this equation now becomes x prime squared times this is root 2 over 2 squared, which is just, 2 4ths or 1 half.
03:56
So we have x prime squared over 2 minus y prime squared over 2.
04:04
And then we have plus x prime y prime times cosine squared of pi over 4.
04:12
This is 1 1⁄2 minus 1 half is 0.
04:22
So plus 0 times x prime y prime and then plus 1 equals 0.
04:30
So we have that y prime squared over 2 minus x prime squared over 2 is equal to 1.
05:00
And so not only have we eliminated the x, y term in the equation, we've written the equation in standard form.
05:13
In standard form, it's easy to see that this is an hyperbola.
05:24
To sketch this hyperbola, well, you notice that the center is the origin.
05:31
Also, y prime squared is a positive term so that the transverse axis is vertical, and a squared is the denominator of y prime squared, 2.
05:48
So, a is equal to root 2.
05:52
And we know that the vertices lie to distance a from the center along the transverse axis...