Q.When the axes are rotated through an angle $45^{\circ}$, the transformed equation of a curve is $17x^2 - 16xy + 17y^2 = 225$. Find the original equation of the curve.

After rotation through $\pi/6$, the equation becomes $X^2 - Y^2 = a^2$.

With $\theta=\pi/6$: $\cos\theta=\dfrac{\sqrt3}{2}$, $\sin\theta=\dfrac12$, so

$x = \frac{\sqrt3}{2}X - \frac12 Y, \qquad y = \frac12 X + \frac{\sqrt3}{2}Y.$

Compute the needed combinations:

$x^2 - y^2 = \frac12 X^2 - \sqrt3\,XY - \frac12 Y^2,$

$2\sqrt3\,xy = \frac32 X^2 + \sqrt3\,XY - \frac32 Y^2.$

Adding, the $XY$ terms cancel:

$x^2 + 2\sqrt3\,xy - y^2 = 2X^2 - 2Y^2.$ …

Substitute the rotation-of-axes formulas (new coordinates in terms of old) into the transformed equation to recover the original equation.

When axes are rotated through angle $\theta$, the new coordinates $(X,Y)$ of a point relate to the old coordinates $(x,y)$ by

$X = x\cos\theta + y\sin\theta, \qquad Y = -x\sin\theta + y\cos\theta$

Here $\theta = 45^\circ$, so $\cos\theta=\sin\theta = \dfrac{1}{\sqrt2}$:

$X = \frac{x+y}{\sqrt2}, \qquad Y = \frac{y-x}{\sqrt2}$

The transformed equation is $17X^2 - 16XY + 17Y^2 = 225$. Substitute:

$X^2 = \frac{(x+y)^2}{2},\quad Y^2=\frac{(x-y)^2}{2},\quad XY = \frac{(x+y)(y-x)}{2}=\frac{y^2-x^2}{2}$

$17X^2+17Y^2 = \frac{17}{2}\left[(x+y)^2+(x-y)^2\right] = \frac{17}{2}(2x^2+2y^2) = 17x^2+17y^2$

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Substitute the standard axis-rotation formulas $x=X\cos\alpha-Y\sin\alpha$, $y=X\sin\alpha+Y\cos\alpha$ into the given equation and simplify using $\sin^2\alpha+\cos^2\alpha=1$.

Given: $x\cos\alpha+y\sin\alpha=P$.

$x\cos\alpha+y\sin\alpha = (X\cos\alpha-Y\sin\alpha)\cos\alpha + (X\sin\alpha+Y\cos\alpha)\sin\alpha$

$= X\cos^2\alpha - Y\sin\alpha\cos\alpha + X\sin^2\alpha + Y\cos\alpha\sin\alpha$

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Substitute the 45° rotation formulas $x=\frac{X-Y}{\sqrt2},\ y=\frac{X+Y}{\sqrt2}$ into the given equation and simplify.

For rotation through $\theta=\dfrac{\pi}{4}$: $x = X\cos\theta - Y\sin\theta = \dfrac{X-Y}{\sqrt2}$, $y = X\sin\theta+Y\cos\theta = \dfrac{X+Y}{\sqrt2}$.

$x^2=\dfrac{(X-Y)^2}{2},\quad y^2=\dfrac{(X+Y)^2}{2},\quad xy=\dfrac{(X-Y)(X+Y)}{2}=\dfrac{X^2-Y^2}{2}$

$3x^2+3y^2 = \dfrac{3}{2}\left[(X-Y)^2+(X+Y)^2\right] = \dfrac{3}{2}(2X^2+2Y^2) = 3X^2+3Y^2$

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For $ax^2+2hxy+by^2$ under rotation $\theta=\frac{\pi}{6}$ ($a=1,h=\sqrt3,b=-1$), the new coefficients are $A'=2$, $B'=-2$, $H'=0$.

Under a rotation through $\theta$, the coefficients of $ax^2+2hxy+by^2$ transform to:

$A'=a\cos^2\theta+2h\cos\theta\sin\theta+b\sin^2\theta$,

$B'=a\sin^2\theta-2h\sin\theta\cos\theta+b\cos^2\theta$,

$H'=(b-a)\sin\theta\cos\theta+h(\cos^2\theta-\sin^2\theta)$.

Here $a=1$, $h=\sqrt3$, $b=-1$, and $\theta=\dfrac{\pi}{6}$: $\cos^2\theta=\tfrac34$, $\sin^2\theta=\tfrac14$, $\sin\theta\cos\theta=\tfrac{\sqrt3}{4}$, $\cos^2\theta-\sin^2\theta=\tfrac12$.

$A'=1\cdot\tfrac34+2\sqrt3\cdot\tfrac{\sqrt3}{4}-1\cdot\tfrac14=\tfrac34+\tfrac64-\tfrac14=2$. …

Reverse the axis-rotation transformation by expressing the new (rotated) coordinates in terms of the original coordinates, then substitute into the given transformed equation.

When axes are rotated through angle $\theta$, a point's new coordinates $(X,Y)$ relate to its original coordinates $(x,y)$ by

$X = x\cos\theta + y\sin\theta, \qquad Y = -x\sin\theta + y\cos\theta$

Here $\theta = 45^\circ$, so $\cos\theta=\sin\theta=\dfrac{1}{\sqrt2}$:

$X = \dfrac{x+y}{\sqrt2}, \qquad Y = \dfrac{y-x}{\sqrt2}$

The transformed equation is $17X^2 - 16XY + 17Y^2 = 225$. Compute each term:

$X^2 = \dfrac{(x+y)^2}{2}, \qquad Y^2 = \dfrac{(y-x)^2}{2}, \qquad XY = \dfrac{(x+y)(y-x)}{2} = \dfrac{y^2-x^2}{2}$

Substitute:

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Substitute the rotation-of-axes formulas $x=X\cos\alpha-Y\sin\alpha$, $y=X\sin\alpha+Y\cos\alpha$ into the given equation and simplify using $\sin^2\alpha+\cos^2\alpha=1$.

When axes are rotated through angle $\alpha$ (origin unchanged), the old coordinates $(x,y)$ relate to new coordinates $(X,Y)$ by:

$x = X\cos\alpha - Y\sin\alpha, \qquad y = X\sin\alpha + Y\cos\alpha$

Substitute into $x\cos\alpha+y\sin\alpha=p$:

$(X\cos\alpha-Y\sin\alpha)\cos\alpha + (X\sin\alpha+Y\cos\alpha)\sin\alpha = p$

$X\cos^{2}\alpha - Y\sin\alpha\cos\alpha + X\sin^{2}\alpha + Y\sin\alpha\cos\alpha = p$

The $Y\sin\alpha\cos\alpha$ terms cancel, and $\cos^2\alpha+\sin^2\alpha=1$:

$X(\cos^{2}\alpha+\sin^{2}\alpha) = p$

$X = p$

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Substituting the rotation formulas for $\theta=\pi/6$ into $x^2+2\sqrt3\,xy-y^2=2a^2$ and simplifying gives $X^2-Y^2=a^2$.

Concept: Rotation of axes

When axes are rotated through angle $\theta$, the old coordinates relate to the new ones by $x = X\cos\theta - Y\sin\theta$, $y = X\sin\theta + Y\cos\theta$.

Step 1: Substitute $\theta=\pi/6$ ($\cos\theta=\frac{\sqrt3}{2}$, $\sin\theta=\frac12$)

$x = \dfrac{\sqrt3}{2}X - \dfrac12 Y$, \quad $y = \dfrac12 X + \dfrac{\sqrt3}{2}Y$

Step 2: Compute $x^2$, $y^2$, $xy$

$x^2 = \dfrac34X^2 - \dfrac{\sqrt3}{2}XY + \dfrac14Y^2$

$y^2 = \dfrac14X^2 + \dfrac{\sqrt3}{2}XY + \dfrac34Y^2$

$xy = \dfrac{\sqrt3}{4}X^2 + \dfrac12XY - \dfrac{\sqrt3}{4}Y^2$

Step 3: Substitute into $x^2+2\sqrt3\,xy-y^2$

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