What does the equation `2x^2+4x y-5y^2+20 x-22 y-14=0` become when referred to the rectangular axes through the point `(-2,-3)` , the new axes being inclined at an angle at `45^0` with the old axes?

Explanation

Let O' be `(-2,-3)`. Since the axes are rotated about O' by an angle `45^@` in the anticlockwise direction, let `(x',y')` be the new coordinates with respect to new axes and `(x,y)` be the coordinates with respect to the old axes. Then, we have
`x=-2+x'cos45^@-y'sin45^@=-2+((x'-y')/(sqrt2))`
`x=-3+x'sin45^@+y'cos45^@=-2+((x'-y')/(sqrt2))`
The new equation will be
`2{-2+((x'-y')/(sqrt2))}^2+4{-2+((x'-y')/(sqrt2))}{-3+((x'-y')/(sqrt2))}`
`+5{-3+((x'-y')/(sqrt2))}+20{-2+((x'-y')/(sqrt2))}`
`-22{-3+((x'-y')/(sqrt2))}-14=0`
or `x'^2-14x'y'-7y'2-2=0`
Hence, the new equation of the curve is `x^2-14xy -7y^2-2=0`