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

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Question

(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}$,the transformation equa

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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}$,the transformation equations are:$x = x_1 \cos \frac{\pi}{4} - y_1 \sin \frac{\pi}{4} = \frac{x_1-y_1}{\sqrt{2}}$$y = x_1 \sin \frac{\pi}{4} + y_1 \cos \frac{\pi}{4} = \frac{x_1+y_1}{\sqrt{2}}$Substituting these into $(i)$:$\left(\frac{x_1-y_1}{\sqrt{2}}ight)^2 + 6\left(\frac{x_1-y_1}{\sqrt{2}}ight)\left(\frac{x_1+y_1}{\sqrt{2}}ight) + 8\left(\frac{x_1+y_1}{\sqrt{2}}ight)^2 = 10$$\frac{x_1^2+y_1^2-2x_1y_1}{2} + \frac{6(x_1^2-y_1^2)}{2} + \frac{8(x_1^2+y_1^2+2x_1y_1)}{2} = 10$Multiplying by $2$:$(x_1^2+y_1^2-2x_1y_1) + (6x_1^2-6y_1^2) + (8x_1^2+8y_1^2+16x_1y_1) = 20$$(1+6+8)x_1^2 + (1-6+8)y_1^2 + (-2+16)x_1y_1 = 20$$15x_1^2 + 3y_1^2 + 14x_1y_1 = 20$Thus,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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