00:01
Hello, we consider the equation xy is equal to zero.
00:07
So we compare this with the form a x squared plus bxy plus c y squared plus dx plus ey plus a plus a plus a plus a plus a plus a is equal to zero we can conclude then that a is equal to c which is equal to zero and b is equal to one so then finding the value of co -tangent of two theta it's going to be equal to a minus c over b we get that cotangent of 2 theta is going to be equal to just 0 minus 0 over 1.
00:40
So just 0.
00:44
And then we get that 2 theta is then equal to the inverse co -tangent of 0.
00:51
So therefore we have that theta is going to be equal to just one -half times pi over 2.
00:58
So therefore theta is equal to pi over four.
01:06
Then using x is equal to x prime, cosine theta minus y prime, sine theta, for theta being equal to sine of pi over four, we get that x is going to be equal to, so x prime of one over, so x prime times one over square root of two, minus y prime times one over square root of two, giving us x prime minus y prime, all divided by square root of two.
01:45
And then for y, we use y is equal to x prime sine of theta plus y prime cosine of theta.
01:53
So we get that y is going to be equal here to x prime plus y prime over square root of two.
02:06
And then we can put these values in we get 0 is equal to x prime minus y prime over square root of two times x prime plus y prime over square to two plus one and then we compare with the form y prime minus k squared over a squared minus x squared or x prime minus h squared over b squared equal to negative one we conclude then that h and k are equal to zero and a and b are equal to square root of two so then we're going to be sketching the equation here, y prime squared over the square to 2 squared minus x prime squared over square to 2 squared 2 squared 2 squared 2 to 1.
03:12
So therefore we then have our graph...