8 Most SAQ’s of Transformation of Axes Chapter in Inter 1st Year Maths-1B (TS/AP)

SAQ-1 : When the origin is shifted to (-1,2) by the translation of axes, find the transformed equation of x² + y² + 2x – 4y + 1 = 0

Given the original equation is

$$x^2 + y^2 + 2x – 4y + 1 = 0$$

We take new origin (h,k) = (-1,2) then

$$x = X + h$$

$$x = X – 1$$

$$y = Y + k$$

$$y = Y + 2$$

From (1) the transformed equation is

$$(X – 1)^2 + (Y + 2)^2 + 2(X – 1) – 4(Y + 2) + 1 = 0$$

$$(X^2 + 1 – 2X) + (Y^2 + 4 + 4Y) + 2X – 2 – 4Y – 8 + 1 = 0$$

$$X^2 + Y^2 – 4 = 0$$

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SAQ-2 : When the origin is shifted to (-1,2) by the translation of axes find the transformed equation of 2x² + y² – 4x + 4y = 0

Given the original equation is

$$2x^2 + y^2 – 4x + 4y = 0$$

We take a new origin (h, k) = (-1, 2) then

$$x = X + h$$

$$x = X – 1$$

$$y = Y + k$$

$$y = Y + 2$$

From (1) the transformed equation is

$$2(X – 1)^2 + (Y + 2)^2 – 4(X – 1) + 4(Y + 2) = 0$$

$$2(X^2 – 2X + 1) + (Y^2 + 4Y + 4) – 4X + 4 + 4Y + 8 = 0$$

$$2X^2 – 4X + 2 + Y^2 + 4Y + 4 – 4X + 4 + 4Y + 8 = 0$$

$$2X^2 + Y^2 – 8X + 8Y + 18 = 0$$

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SAQ-3 : When the origin is shifted to the point (2,3) the transformed equation of a curve is x² + 3xy – 2y² + 17z – 7y – 11 = 0 find the original equation of the curve

Given the transformed equation is

$$X^2 + 3XY – 2Y^2 + 17X – 7Y – 11 = 0.$$

When we take the origin as (h, k) = (2, 3) then

$$X = x – h \Rightarrow X = x – 2$$

$$Y = y – k \Rightarrow Y = y – 3.$$

Substituting these values into the equation gives:

$$(x – 2)^2 + 3(x – 2)(y – 3) – 2(y – 3)^2 + 17(x – 2) – 7(y – 3) – 11 = 0.$$

$$x^2 + 4 – 4x + 3(xy – 3x – 2y + 6) – 2(y^2 + 9 – 6y) + 17x – 34 – 7y + 21 – 11 = 0.$$

$$x^2 + 3xy – 2y^2 + 4x – y – 20 = 0.$$

Hence, the original equation is

$$x^2 + 3xy – 2y^2 + 4x – y – 20 = 0.$$

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SAQ-4 : Find the transformed equation of x² + 2√3xy – y² = 2a² when the axes are rotated through an angle π/6

Given equation is

$$x^2 + 2\sqrt{3}xy – y^2 = 2a^2\quad…(1)$$

Angle of rotation $$\theta = \pi/6 = 300$$ then

$$x = X\cos\theta – Y\sin\theta$$

$$\Rightarrow x = X\cos 300 – Y\sin 300$$

$$= X\left(\sqrt{3}/2\right) – Y\left(1/2\right)$$

$$x = \sqrt{3}X – Y/2$$

$$y = Y \cos\theta + X \sin\theta$$

$$y = Y \cos 300 + X \sin300$$

$$= Y\left(\sqrt{3}/2\right) + X\left(1/2\right)$$

$$y = \sqrt{3}Y + X/2$$

From (1) transformed equation is

$$(\sqrt{3}X – Y/2)^2 + 2\sqrt{3}(\sqrt{3}X – Y/2)(\sqrt{3}Y + X/2) – (\sqrt{3}Y + X/2)^2 = 2a^2$$

$$(\sqrt{3}X – Y)^2 + 2\sqrt{3}(\sqrt{3}X – Y)(\sqrt{3}Y + X) – (\sqrt{3}Y + X)^2/4 = 2a^2$$

$$(3X^2 + Y^2 – 2\sqrt{3}XY) + 2\sqrt{3}(3XY + \sqrt{3}X^2 – \sqrt{3}Y^2 – XY) – (3Y^2 + X^2 + 2\sqrt{3}XY) = 4(2a^2)$$

$$3X^2 + Y^2 – 2\sqrt{3}XY + 6\sqrt{3}XY + 6X^2 – 6Y^2 – 2\sqrt{3}XY – 3Y^2 – X^2 – 2\sqrt{3}XY = 8a^2$$

$$8X^2 – 8Y^2 = 8a^2$$

$$8(X^2 – Y^2) = 8a^2$$

$$X^2 – Y^2 = a^2$$

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SAQ-5 : Find the transformed equation of 3x² + 10xy + 3y² = 9 when the axes are rotated an angle π/4

Given equation original is

$$3x^2 + 10xy + 3y^2 = 9\quad…(1)$$

Angle of rotation $$\theta = \pi/4 = 45^\circ$$

$$x = X\cos\theta – Y\sin\theta$$

$$\Rightarrow x = X \cos 45^\circ – Y \sin 45^\circ$$

$$= X\left(\frac{1}{\sqrt{2}}\right) – Y\left(\frac{1}{\sqrt{2}}\right)$$

$$\Rightarrow x = \frac{X – Y}{\sqrt{2}}$$

$$y = Y \cos\theta + X \sin\theta$$

$$\Rightarrow y = Y \cos 45^\circ + X \sin 45^\circ$$

$$= Y\left(\frac{1}{\sqrt{2}}\right) + X\left(\frac{1}{\sqrt{2}}\right)$$

$$\Rightarrow y = \frac{X + Y}{\sqrt{2}}$$

From (1) transformed equation is

$$3\left(\frac{X – Y}{\sqrt{2}}\right)^2 + 10\left(\frac{X – Y}{\sqrt{2}}\right)\left(\frac{X + Y}{\sqrt{2}}\right) + 3\left(\frac{X + Y}{\sqrt{2}}\right)^2 = 9$$

$$\Rightarrow 3(X^2 + Y^2 – 2XY/\sqrt{2}) + 10(X^2 – Y^2/\sqrt{2}) + 3(X^2 + Y^2 + 2XY/\sqrt{2}) = 9$$

$$\Rightarrow 3X^2 + 3Y^2 – 6XY/\sqrt{2} + 10X^2 – 10Y^2/\sqrt{2} + 3X^2 + 3Y^2 + 6XY/\sqrt{2} = 9$$

$$\Rightarrow 16X^2 – 4Y^2 = 2(9)$$

$$\Rightarrow 2(8X^2 – 2Y^2) = 2(9)$$

$$\Rightarrow 8X^2 – 2Y^2 = 9$$

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SAQ-6 : When the axes are rotated through an angle α, find the transformed equation of x cosα + ysinα = p

Given original equation is

$$x \cos\alpha + y \sin\alpha = p\quad…(1)$$

Angle of rotation $$\theta = \alpha$$ then

$$x = X\cos\theta – Y\sin\theta$$

$$\Rightarrow x = X\cos\alpha – Y\sin\alpha$$

$$y = Y\cos\theta + X\sin\theta$$

$$\Rightarrow y = Y \cos\alpha + X \sin\alpha$$

From (1) transformed equation is

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

$$\Rightarrow X\cos^2\alpha – Y\sin\alpha\cos\alpha + Y\cos\alpha\sin\alpha + X\sin^2\alpha = p$$

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

$$\Rightarrow X(1) = p$$

$$\Rightarrow X = p$$

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SAQ-7 : When the axes are rotated through an angle 45⁰ the transformed equation of a curve is 17x² – 16xy + 17y² = 225. Find the original equation of the curve

Given transformed equation is taken as

$$17X^2 – 16XY + 17Y^2 = 225\quad…(1)$$

Angle of rotation $$\theta = 45^\circ$$ then

$$X = x\cos\theta + y\sin\theta = x\cos 45^\circ + y\sin 45^\circ$$

$$= x\left(\frac{1}{\sqrt{2}}\right) + y\left(\frac{1}{\sqrt{2}}\right)$$

$$\Rightarrow X = \frac{x + y}{\sqrt{2}}$$

$$Y = y\cos\theta – x\sin\theta = y\cos 45^\circ – x\sin 45^\circ$$

$$= y\left(\frac{1}{\sqrt{2}}\right) – x\left(\frac{1}{\sqrt{2}}\right)$$

$$\Rightarrow Y = \frac{y – x}{\sqrt{2}}$$

From (1) original equation is

$$$$17\left(\frac{x + y}{\sqrt{2}}\right)^2 – 16\left(\frac{x + y}{\sqrt{2}}\right)\left(\frac{y – x}{\sqrt{2}}\right) + 17\left(\frac{y – x}{\sqrt{2}}\right)^2 = 225$$

$$\Rightarrow 17\left(x^2 + y^2 + \frac{2xy}{\sqrt{2}}\right) – 16\left(y^2 – \frac{x^2}{\sqrt{2}}\right) + 17\left(y^2 + x^2 – \frac{2xy}{\sqrt{2}}\right) = 225$$

$$\Rightarrow 17x^2 + 17y^2 + 34xy – 16y^2 + 16x^2 + 17x^2 + 17y^2 – 34xy/2 = 225$$

$$\Rightarrow 50x^2 + 18y^2 = 2(225)$$

$$\Rightarrow 2(25x^2 + 9y^2) = 2(225)$$

$$\Rightarrow 25x^2 + 9y^2 = 225$$

Hence, the required original equation is $$25x^2 + 9y^2 = 225$$

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SAQ-8 : Prove that the angle of rotation of the axes to eliminate xy term from the equation ax² + 2hxy + by2 = 0 is 1/2Tan^-1(2h/a-b) where a ≠ b and π/4 if a = b

Given the original equation is

$$ax^2 + 2hxy + by^2 = 0\quad…(1)$$

Angle of rotation is $$\theta$$ then

$$x = X\cos\theta – Y\sin\theta$$

$$y = Y\cos\theta + X\sin\theta$$

From (1), the transformed equation is

$$a(X\cos\theta – Y\sin\theta)^2 + 2h(X\cos\theta – Y\sin\theta)(Y\cos\theta + X\sin\theta) + b(Y\cos\theta + X\sin\theta)^2 = 0$$

$$\Rightarrow a(X^2\cos^2\theta + Y^2\sin^2\theta – 2XY\cos\theta\sin\theta) + 2h(XY\cos^2\theta + X^2\cos\theta\sin\theta – Y^2\sin\theta\cos\theta – YX\sin^2\theta) + b(Y^2\cos^2\theta + X^2\sin^2\theta + 2XY\sin\theta\cos\theta) = 0\quad…(2)$$

Equating coefficients of XY to zero in (2), we get

$$-2a\cos\theta\sin\theta + 2h\cos^2\theta – 2h\sin^2\theta + 2b\sin\theta\cos\theta = 0$$

$$\Rightarrow -[2a\cos\theta\sin\theta – 2h\cos^2\theta + 2h\sin^2\theta – 2b\sin\theta\cos\theta] = 0$$

$$\Rightarrow 2a\cos\theta\sin\theta – 2h(\cos^2\theta – \sin^2\theta) – 2b\sin\theta\cos\theta = 0$$

$$\Rightarrow 2\sin\theta\cos\theta (a-b) = 2h(\cos^2\theta – \sin^2\theta)$$

$$\Rightarrow \sin 2\theta(a-b) = 2h\cos 2\theta$$

$$\Rightarrow \frac{\sin 2\theta}{\cos 2\theta} = \frac{2h}{a-b}, a \neq b$$

$$\Rightarrow \tan 2\theta = \frac{2h}{a-b}$$

$$\Rightarrow 2\theta = \tan^{-1}\left(\frac{2h}{a-b}\right)$$

$$\Rightarrow \theta = \frac{1}{2} \tan^{-1}\left(\frac{2h}{a-b}\right)$$

If a = b then $$\theta = \frac{1}{2}\left(\frac{\pi}{2}\right) = \frac{\pi}{4}$$