Let C be a curve ax^2+2hxy+by^2+2gx+2fy+c=0 i | vedclass

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(A) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. $\ldots(i)$Rotating the axes by $	heta = \frac{\pi}{4}$ implies $x

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(A) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$. $\ldots(i)$Rotating the axes by $	heta = \frac{\pi}{4}$ implies $x = X\cos	heta - Y\sin	heta = \frac{X-Y}{\sqrt{2}}$ and $y = X\sin	heta + Y\cos	heta = \frac{X+Y}{\sqrt{2}}$.Substituting these into $(i)$ gives the transformed equation:$\frac{a}{2}(X-Y)^2 + h(X^2-Y^2) + \frac{b}{2}(X+Y)^2 + \sqrt{2}g(X-Y) + \sqrt{2}f(X+Y) + c = 0$.Expanding and grouping terms:$(\frac{a+b}{2}+h)X^2 + (b-a)XY + (\frac{a+b}{2}-h)Y^2 + \sqrt{2}(g+f)X + \sqrt{2}(f-g)Y + c = 0$.Comparing with $Y^2+XY-X=0$,we get:$1) \frac{a+b}{2}+h = 0$$2) b-a = 1$$3) \frac{a+b}{2}-h = 1$$4) \sqrt{2}(g+f) = -1$$5) f-g = 0$From $(1)$ and $(3)$,$a+b=1$ and $h=-\frac{1}{2}$.From $(2)$,$b-a=1$. Solving $a+b=1$ and $b-a=1$ gives $b=1, a=0$.From $(4)$ and $(5)$,$f=g=-\frac{1}{2\sqrt{2}}$.Now,$(h^2-ab)-2gf = ((-\frac{1}{2})^2 - (0)(1)) - 2(-\frac{1}{2\sqrt{2}})(-\frac{1}{2\sqrt{2}}) = \frac{1}{4} - 2(\frac{1}{8}) = \frac{1}{4} - \frac{1}{4} = 0$.

Let $C$ be a curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ in a Cartesian plane. By rotating the coordinate axes through an angle $\frac{\pi}{4}$ in the positive direction,if the transformed equation of $C$ is $Y^2+XY-X=0$,then $(h^2-ab)-2gf=$

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