conic problem

[Sonal7]

Here is a nice question. I did the first part and got the values for x and y. But I am not sure how they got the equation?

I understand the formula for ellipse and how to derive it if the ellipse had not been rotated. [MATH]\frac{x^2}{a^2}[/MATH]+[MATH]\frac{y^2}{b^2}[/MATH]. My only confusion is why x is not x+y, have they made a mistake? should it not be x-y as per the new values for x and y. Sorry minor question but I think its important. There is a typing error the y value should be b sin [MATH]\theta[/MATH]?

I conclude there are typing mistakes, the minus should be in the new X value and the plus in the new Y value.

 

[Sonal7]

I am still not sure how they got the equation.

 

[Dr.Peterson]

I'm not sure what you are asking.

I presume the first picture is the problem, and the second is someone's solution to it. But in your question, what are "the values for x and y"? I see no values. (I suppose you must mean the expressions in the vector they obtained.) And what does "why x is not x+y" mean? Of course x is not the same as x+y!

As far as I can see, the solution is not "getting" the equation [MATH]\frac{(x+y)^2}{2a^2} + \frac{(x-y)^2}{2b^2} = 1[/MATH], but taking it as given (as it was in the problem) and checking that it is satisfied by points on the rotated ellipse. They just "compute" the LHS and show that it is 1, by adding and subtracting the expressions for the "new" x and y to get x+y and x-y.