Students must practice these TS Intermediate Maths 1B Solutions Chapter 2 Transformation of Axes Ex 2(a) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1B Transformation of Axes Solutions Exercise 2(a)

I.

Question 1.

When the origin is shifted to (4, – 5) by the translation of axes, find the coordinates of the following points with reference to new axes. (V.S.A.Q.)

(i) (0, 3)

(ii) (-2, 4)

(iii) (4, -5)

Answer:

(i) (0, 3)

New origin = (4, -5) = (h, k)

Old co-ordinates are (0, 3) = (x, y)

x’ = x – h = 0 – 4 = – 4 and y’ = y – k = 3 + 5 = 8

∴ New coordinates = (- 4, 8)

(ii) (- 2, 4)

New origin = (4, -5) = (h, k)

Old coordinates are (- 2, 4) = (x, y)

x’ = x – h = – 2 – 4 = – 6

y’ = y – k = 4 + 5 = 9

∴ New coordinates = (- 6, 9)

(iii) (4, – 5)

New origin = (4, -5) = (h, k)

Old coordinates are (4, -5) = (x, y)

x’ = x – h = 4 – 4 = 0

y’ = y – k = – 5 + 5 = 0

∴ New coordinates = (0, 0)

Question 2.

The origin is shifted to (2,3) by the translation of axes. If the co-ordinates of a point P change as follows, find the co-ordinates of P in the original system. (V.S.A.Q.)

(i) (4, 5)

(ii) (-4, 3)

(iii)(0, 0)

Answer:

(i) (4, 5)

New coordinates (x’,y’) = (4, 5)

origin (h, k) = (2, 3)

Old coordinates of P are x = x’ + h = 4 + 2 = 6

y = y’ + k = 5 + 3 = 8

∴ Old coordinates = (6, 8)

(ii) (- 4, 3)

New coordinates (x’, y’) = (-4, 3)

origin (h, k) = (2, 3)

Old coordinates of P are x = x’ + h = – 4 + 2 = – 2

y = y’ + k = 3 + 3 = 6

∴ Old coordinates = (- 2, 6)

(iii) (0, 0)

New coordinates (x’,y’) = (0, 0)

origin(h, k) = (2, 3)

Old coordinates of P are x = x’ + h = 0 + 2 = 2

y = y’ + k = 0 + 3 = 3

∴ Old co-ordinates = (2, 3)

Question 3.

Find the point to which the origin is to be shifted so that the point (3,0) may change to (2, -3). (V.S.A.Q.)

Answer:

(x, y) = (3, 0)

(x’,y’) = (2,-3)

Let ( h, k ) be the shifted origin,

h = x – x’ = 3 – 2 = 1

k = y – y’ = 0 + 3 = 3

∴ (h, k) = (1, 3)

Question 4.

When the origin is shifted to (-1, 2) by the translation of axes, find the transformed equations of the following.

(i) x^{2} + y^{2} + 2x – 4y + 1 = 0

(ii) 2x^{2} + y^{2} – 4x + 4y = 0 (V.S.A.Q.)

Answer:

(i) x^{2} + y^{2} + 2x – 4y + 1 = 0

The given equation is

x^{2} + y^{2} + 2x – 4y + 1 = 0

origin is shifted to (-1, 2)

∴ h = – 1, k = 2

Transformed equations are

x = x’ + h, y = y’+ k

⇒ x = x’ – 1, y = y’ + 2

∴ The new equation is

(x’ – 1)^{2} + (y’ + 2)^{2} + 2( x’ – 1) – 4(y’ + 2) + 1 = 0

x’^{2} + 1 – 2x’ + y’^{2} + 4y’ + 4 + 2x ‘ – 2 – 4y’- 8 + 1 = 0

⇒ x’^{2} + y’^{2} – 4 = 0

(ii) 2x^{2} + y^{2} – 4x + 4y = 0

The given equation is

2x^{2} + y^{2} – 4x + 4y = 0

By the above transformation we have

x = x’ – 1, y = y’ + 2

∴ 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 + 4 y’ + 8 = 0

⇒ 2 x’^{2} + y’^{2} – 8x’ + 8y’ + 18 = 0

Question 5.

The point to which the origin is shifted and the transformed equations are given below. Find the original equation.

(i) ( 3, – 4 ) ; x^{2} + y^{2} = 4

(ii) (-1, 2 ) ; x^{2} + 2y^{2} + 16 = 0 (V.S.A.Q.)

Answer:

(i) ( 3, – 4 ) ; x^{2} + y^{2} = 4

Shifted origin = (h, k) = ( 3, – 4)

x’ = x – h

= x – 3

y’ = y – k

= y + 4

The given equation is x^{2} + y^{2} = 4

The original equation of 3x’^{2} + y’^{2} = 4 is

(x – 3)^{2} + (y + 4)^{2} = 4

⇒ x^{2} – 6x + 9 + y^{2} + 8y + 16 = 4

⇒ x^{2} + y^{2} – 6x + 8y + 21 = 0

(ii) (-1, 2) ; x^{2} + 2y^{2} + 16 = 0

Given shifted origin = (h, k) = (- 1, 2)

x’=x + h

= x + 1

y’ = y + k

= y – 2

The original equation of x’^{2} +2 y’^{2} + 16 = 0 is

( x + 1)^{2} + 2 (y – 2)^{2} + 16 = 0

⇒ x^{2} + 2x + 1 + 2 ( y^{2} – 4y + 4) + 16 = 0

⇒ x^{2} + 2y^{2} + 2x – 8y + 25 = 0

Question 6.

The point to which the origin is shifted and the transformed equations are given below.

Find the original equation.

(i) (3, – 4); x^{2} + y^{2} = 4

(ii) (- 1, 2); x^{2} + 2y^{2} + 16 = O (V.S.A.Q.)

Answer:

(3, – 4); x^{2} + y^{2} = 4

Shifted origin (h, k) = (3, – 4)

x’ = x – h

= x – 3

y’ = y – k

= y + 4

The given equation is x^{2} + y^{2} = 4

The original equation of 3x’^{2} + y’^{2} = 4 is

(x – 3)^{2} + (y + 4)^{2} = 4

⇒ x^{2} – 6x + 9 + y^{2} + 8y + 16 = 4

⇒ x^{2} + y^{2} – 6x + 8y + 21 = 0

(ii) (- 1, 2) ; x^{2} + 2y^{2} + 16 = 0

Given shifted origin = (h, k) = (- 1, 2)

x’ = x + h

= x + 1

y’ = y + k

= y – 2

The origin equation of x’^{2} + 2y’^{2} + 16 = 0 is

(x + 1)^{2} + 2(y – 2)^{2} + 16 = 0

⇒ x^{2} + 2x + 1 + 2 (y^{2} – 4y + 4) + 16 = 0

⇒ x^{2} + 2y^{2} + 2x – 8y + 25 = 0

Question 7.

Find the point to which the origin is to be shifted so as to remove the first degree terms from the equation 4x^{2} + 9y^{2} – 8x + 36y + 4 =0. (V.S.A.Q.)

Answer:

The given equation is

4x^{2} + 9y^{2} – 8x + 36y + 4 = 0

To make first degree terms missing in the equation origin should be shifted to \(\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)\)

Comparing with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 we have

a = 4, b = 9, h = 0, g = – 4, f = 18, c = 4

Question 8.

When the axes are rotated through an angle 30°, find the new coordinates of the following points. (V.S.A.Q.)

(i) (0, 5)

(ii) (- 2, 4)

(iii) (0, 0)

Answer:

(i) ( 0, 5 )

Given θ = 30° and (x, y) = (0, 5)

We have x’ = x cos θ + y sin θ

= 0 (cos 30°) + 5 ( sin 30°) = 5\(\left(\frac{1}{2}\right)\) = \(\frac{5}{2}\)

y’ = – x sin θ + y cos θ

= – 0 (sin 30°) + 5 cos 30° = \(\frac{5 \sqrt{3}}{2}\)

∴ New coordinates = \(\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)\)

(ii) (-2, 4)

θ = 30° and (x, y) = (-2, 4)

We have x’= x cos θ + y sin θ

= – 2 cos 30° + 4 sin 30°

= – 2 \(\left(\frac{\sqrt{3}}{2}\right)\) + 4\(\left(\frac{1}{2}\right)\) = – √3 + 2

y’ = – x sin θ + y cos θ

= – (- 2) sin 30° + 4 cos 30°

= 2 sin 30° + 4 cos 30°

= 2 \(\left(\frac{1}{2}\right)\) + 4 \(\left(\frac{\sqrt{3}}{2}\right)\) = 1 + 2√3

∴ New coordinates = (- √3 + 2, 1 + 2√3)

(iii) (0, 0)

θ = 30° and (x, y) = (0, 0)

x’= x cos θ + y sin θ

= 0 cos 30° + 0 sin 30° = 0

y’ = – x sin 0 + y cos 0

= – (0) sin 30° + 0 cos (30°) = 0

∴ New coordinates = (0, 0)

Question 9.

When the axes are rotated through an angle 60°, the new coordinates of three points are the following

(i) (3, 4)

(ii) (- 7, 2)

(iii) (2, 0)

Find their original coordinates. (V.S.A.Q.)

Answer:

Given θ = 60°, (x’, y’) = (3, 4)

We have x = x’ cos θ – y’ sin θ

= 3 cos 60° – 4 sin 60°

(ii) Given θ = 60° and (x’, y’) = (-7, 2)

We have x = x’ cos θ – y’ sin θ

= (- 7) cos 60° – 2 sin 60°

(iii) Given θ = 60°; (x’, y’) = (2, 0)

We have x = x’ cos θ – y’ sin θ

= 2 cos 60° – 0 sin 60°

= 2\(\left(\frac{1}{2}\right)\) – 0 \(\left(\frac{\sqrt{3}}{2}\right)\) = 1

Also y = x’ sin θ + y’ cos θ

= 2 sin 60° + 0 cos 60° = √3

= 2\(\left(\frac{\sqrt{3}}{2}\right)\) + 0\(\left(\frac{1}{2}\right)\) = √3

∴ Original coordinates of P = (1, √3)

Question 10

Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x^{2} + 4xy + y^{2} – 2x + 2y – 6 = 0. (V.S.A.Q.) (June 2004)

Answer:

Comparing the given equation with

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

We have a = 1, 2h = 4 ⇒ h = 2,

b = 1, 2g = – 2 ⇒ g = – 1

2f = 2 ⇒ f = 1, c = – 6

Let θ be the angle of rotation of axes, then

II.

Question 1.

When the origin is shifted to the point (2, 3), the transformed equation of a curve is x^{2} + 3xy – 2y^{2} + 17x – 7y – 11 = 0. Find the original equation of the curve. (S.A.Q.) (March 2011)

Answer:

Transformed equations are

Transformed equation is

x’^{2} + 3x’y’ – 2 y’^{2} + 17x’ – 7y’ – 11 = 0

Original equation is

(x – 2)^{2} + 3 (x – 2) (y – 3) – 2 (y – 3)^{2} – 7 (y – 3) – 11 = 0

⇒ x^{2} – 4x + 4 + 3 (xy – 2y – 3x + 6) – 2 (y^{2} – 6y + 9) – 7y + 21 – 11 = 0

⇒ x^{2} + 3xy – 2y^{2} + 4x – y – 20 = 0

∴ The original equation of the curve is

x^{2} + 3xy – 2y^{2} + 4x – y – 20 = 0

Question 2.

When the axes are rotated through an angle 45°, the transformed equation of a curve is 17x^{2} – 16xy + 17y^{2} = 225. Find the original equation of the curve. (S.A.Q.) (May 2012)

Answer:

Angle of rotation is θ = 45°

x’ = x cos θ + y sin θ

= x cos 45° + y sin 45°

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

y’ = – x sin θ + y cos θ

= – x sin 45° + y cos 45°

= – \(\frac{x+y}{\sqrt{2}}\)

The original equation of the curve

⇒ 17x^{2} – 16x’y + 17y’^{2} = 225 is

⇒ 17x^{2} + 17y^{2} + 34xy – 16y^{2} + 16x^{2} + 17x^{2} + 17y^{2} – 34xy = 450

⇒ 50x^{2} + 18y^{2} = 450

⇒ 25x^{2} + 9y^{2} = 225

Question 3.

When the axes are rotated through an angle a, find the transformed equation of x cos α + y sin α = p. (S.A.Q.) (Mar. ’14, May 2007)

Answer:

Given equation is x cos α + y sin α = p and axes are rotated through an angle α.

x = x’ cos α – y’ sin α

y = x’ sin α + y’ cos α

The given equation transformed to

(x’ cos α – y’ sin α) cos α + ( x’ sin α + y’ cos α) sin α = p

⇒ x’ (cos^{2}α + sin^{2}α) = p

⇒ x’ = p

∴ Transformed equation is x = p.

Question 4.

When the axes are rotated through an angle \(\frac{\pi}{6}\), find the transformed equation of x^{2} + 2 √3 xy – y^{2} = 2a^{2}. (S.A.Q.) (March 2012, ’07, 04; May 2006 )

Answer:

⇒ 3x^{2} – 2√3 + y^{2} + 6x^{2} – 2√3 xy + 6√3 xy – 6y^{2} – x^{2} – 2√3 xy – 3y^{2} = 8a^{2}

⇒ 8x^{2} – 8y^{2} = 8a^{2}

⇒ x^{2} – y^{2} = a^{2}

∴ Required transformed equation is x^{2} – y^{2} = a^{2}

Question 5.

When the axes are rotated through an angle \(\frac{\pi}{4}\), find the transformed equation of

3x^{2} + 10xy + 3y^{2} = 9. (S.A.Q.) (May’14, ’11)

Answer:

Given equation is

3x^{2} + 10xy + 3y^{2} = 9 …………… (1)

and angle of rotation θ = \(\frac{\pi}{4}\)

Let (X ,Y) be the new coordinates of (x, y) then x = X cos θ – Y sin θ

⇒ 3X^{2} – 6XY + 3Y^{2} + 10X^{2} – 10Y^{2} + 3X^{2} + 6XY + 3Y^{2} = 18

⇒ 16X^{2} – 4Y^{2} – 18 = 0

⇒ 8X^{2} – 2Y^{2} = 9

∴ The transformed equation of the given equation is

8X^{2} – 2Y^{2} = 9