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Question

(C) The rotation of coordinate axes by an angle $	heta$ is given by the transformation: $x = X \cos 	heta - Y \sin 	heta$ $y = X \sin 	heta + Y \c

Vedclass · Accepted Answer

$(p+q+r-15)^2=t$

Vedclass · Answer

(C) The rotation of coordinate axes by an angle $	heta$ is given by the transformation: $x = X \cos 	heta - Y \sin 	heta$ $y = X \sin 	heta + Y \cos 	heta$ Given $	heta = \frac{\pi}{4} = 45^{\circ}$,we have $\cos 	heta = \sin 	heta = \frac{1}{\sqrt{2}}$. Thus,$x = \frac{X-Y}{\sqrt{2}}$ and $y = \frac{X+Y}{\sqrt{2}}$. Substituting these into the equation $49x^2 + 25y^2 = 1225$: $49 \left( \frac{X-Y}{\sqrt{2}} ight)^2 + 25 \left( \frac{X+Y}{\sqrt{2}} ight)^2 = 1225$ $\frac{49}{2} (X^2 + Y^2 - 2XY) + \frac{25}{2} (X^2 + Y^2 + 2XY) = 1225$ Multiplying by $2$: $49(X^2 + Y^2 - 2XY) + 25(X^2 + Y^2 + 2XY) = 2450$ $(49+25)X^2 + (-98+50)XY + (49+25)Y^2 = 2450$ $74X^2 - 48XY + 74Y^2 = 2450$ Dividing by $2$: $37X^2 - 24XY + 37Y^2 = 1225$ Comparing with $px^2 + qxy + ry^2 = t$,we get $p=37, q=-24, r=37, t=1225$. Checking option $C$: $(p+q+r-15)^2 = (37 - 24 + 37 - 15)^2 = (35)^2 = 1225 = t$.

When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi}{4}$,if the equation $49x^2+25y^2=1225$ is transformed to $px^2+qxy+ry^2=t$ and the $G.C.D$ of $p, q, r, t$ is $1$,then:

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