Transformation Of Axes (SAQs)
Maths-1B | 2. Transformation Of Axes – SAQs:
Welcome to SAQs in Chapter 2: Transformation Of Axes. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
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
$$(X – 1)^2 + (Y + 2)^2 + 2(X – 1) – 4(Y + 2) + 1 = 0$$
$$(X^2 + 1 – 2X) + (Y^2 + 4 + 4Y) + 2X – 2 – 4Y – 8 + 1 = 0$$
$$X^2 + Y^2 – 4 = 0$$
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
$$2(X – 1)^2 + (Y + 2)^2 – 4(X – 1) + 4(Y + 2) = 0$$
$$2(X^2 – 2X + 1) + (Y^2 + 4Y + 4) – 4X + 4 + 4Y + 8 = 0$$
$$2X^2 – 4X + 2 + Y^2 + 4Y + 4 – 4X + 4 + 4Y + 8 = 0$$
$$2X^2 + Y^2 – 8X + 8Y + 18 = 0$$
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
$$x^2 + 3xy – 2y^2 + 4x – y – 20 = 0.$$
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
$$(\sqrt{3}X – Y/2)^2 + 2\sqrt{3}(\sqrt{3}X – Y/2)(\sqrt{3}Y + X/2) – (\sqrt{3}Y + X/2)^2 = 2a^2$$
$$(\sqrt{3}X – Y)^2 + 2\sqrt{3}(\sqrt{3}X – Y)(\sqrt{3}Y + X) – (\sqrt{3}Y + X)^2/4 = 2a^2$$
$$(3X^2 + Y^2 – 2\sqrt{3}XY) + 2\sqrt{3}(3XY + \sqrt{3}X^2 – \sqrt{3}Y^2 – XY) – (3Y^2 + X^2 + 2\sqrt{3}XY) = 4(2a^2)$$
$$3X^2 + Y^2 – 2\sqrt{3}XY + 6\sqrt{3}XY + 6X^2 – 6Y^2 – 2\sqrt{3}XY – 3Y^2 – X^2 – 2\sqrt{3}XY = 8a^2$$
$$8X^2 – 8Y^2 = 8a^2$$
$$8(X^2 – Y^2) = 8a^2$$
$$X^2 – Y^2 = a^2$$
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
$$3\left(\frac{X – Y}{\sqrt{2}}\right)^2 + 10\left(\frac{X – Y}{\sqrt{2}}\right)\left(\frac{X + Y}{\sqrt{2}}\right) + 3\left(\frac{X + Y}{\sqrt{2}}\right)^2 = 9$$
$$\Rightarrow 3(X^2 + Y^2 – 2XY/\sqrt{2}) + 10(X^2 – Y^2/\sqrt{2}) + 3(X^2 + Y^2 + 2XY/\sqrt{2}) = 9$$
$$\Rightarrow 3X^2 + 3Y^2 – 6XY/\sqrt{2} + 10X^2 – 10Y^2/\sqrt{2} + 3X^2 + 3Y^2 + 6XY/\sqrt{2} = 9$$
$$\Rightarrow 16X^2 – 4Y^2 = 2(9)$$
$$\Rightarrow 2(8X^2 – 2Y^2) = 2(9)$$
$$\Rightarrow 8X^2 – 2Y^2 = 9$$
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
$$(X\cos\alpha – Y\sin\alpha)\cos\alpha + (Y\cos\alpha + X\sin\alpha)\sin\alpha = p$$
$$\Rightarrow X\cos^2\alpha – Y\sin\alpha\cos\alpha + Y\cos\alpha\sin\alpha + X\sin^2\alpha = p$$
$$\Rightarrow X(\cos^2\alpha + \sin^2\alpha) = p$$
$$\Rightarrow X(1) = p$$
$$\Rightarrow X = p$$
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 $$25x^2 + 9y^2 = 225$$
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 $$\theta = \frac{1}{2}\left(\frac{\pi}{2}\right) = \frac{\pi}{4}$$