The transformed equation of x^2+6xy+8y^2=10 w | vedclass

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(C) The given equation is $x^2+6xy+8y^2=10$ $\dots$ $(i)$.Since the axes are rotated through an angle $	heta = \frac{\pi}{4}$,we use the transformati

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$15x^2+14xy+3y^2=20$

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(C) The given equation is $x^2+6xy+8y^2=10$ $\dots$ $(i)$.Since the axes are rotated through an angle $	heta = \frac{\pi}{4}$,we use the transformation formulas:$x = x' \cos 	heta - y' \sin 	heta = \frac{x'-y'}{\sqrt{2}}$$y = x' \sin 	heta + y' \cos 	heta = \frac{x'+y'}{\sqrt{2}}$Substituting these into $(i)$:$\left(\frac{x'-y'}{\sqrt{2}}ight)^2 + 6\left(\frac{x'-y'}{\sqrt{2}}ight)\left(\frac{x'+y'}{\sqrt{2}}ight) + 8\left(\frac{x'+y'}{\sqrt{2}}ight)^2 = 10$$\frac{x'^2+y'^2-2x'y'}{2} + 6\left(\frac{x'^2-y'^2}{2}ight) + 8\left(\frac{x'^2+y'^2+2x'y'}{2}ight) = 10$Multiplying by $2$:$(x'^2+y'^2-2x'y') + 6(x'^2-y'^2) + 8(x'^2+y'^2+2x'y') = 20$$x'^2+y'^2-2x'y' + 6x'^2-6y'^2 + 8x'^2+8y'^2+16x'y' = 20$$(1+6+8)x'^2 + (1-6+8)y'^2 + (-2+16)x'y' = 20$$15x'^2 + 3y'^2 + 14x'y' = 20$Replacing $x', y'$ with $x, y$,the transformed equation is $15x^2+14xy+3y^2=20$.

The transformed equation of $x^2+6xy+8y^2=10$ when the axes are rotated through an angle $\frac{\pi}{4}$ is:

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