Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Transformation of Axes Solutions Exercise 2(a) will help students to clear their doubts quickly.

## Intermediate 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.
i) (0, 3), ii) (-2, 4) iii) (4, -5)
Solution:
i) New origin = (4, -5); h = 4, k = -5
Old co-ordinates are (0, 3)
x = 0, y = 3
x’ = x – h = 0 – 4 = -4
y’=y-k = 3 + 5 = 8
New co-ordinates are (—4, 8)

ii) Old co-ordinates are (-2, 4)
x = -2, y = 4
x’ = x- h = -2 – 4 = -6
y’=y-k=4 + 5 = 9
New co-ordinates are (-6, 9)

iii) Old co-ordinates are (4, -5)
x = 4, y = -5
x’ = x – h = 4- 4 = 0
y = y – k = -5 + 5 = 0
New co-ordinates are (0,0)

Question 2.
The origin is shifted to (2, 3) by the translation of axes. If the coordinates of a point P changes as follows, find the coordinates of P in the original system.
i) (4, 5) ii) (-4, 3), iii) (0, 0)
Solution:
i) New co-ordinates are (4, 5)
x’ = 4, y’ = 5
x = x’ + h = 4 + 2 = 6
y = y’ + k = 5 + 3 = 8
Old co-ordinates are (6, 8)

ii) New co-ordinates are (-4, 3)
x’ = – 4, y’ = 3
x = x’ + h = -4 + 2 = -2
y = y’ + k = 3 + 3 = 6
Old co-ordinates are (-2, 6)

iii) New co-ordinates are (0, 0)
x’ = 0, y’ = 0
x = x’ + h = 0 + 2 = 2
y = y’ + k = 0 + 3 = 3
Old co-ordinates are (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).
Solution:
(x, y) = (3, 0)
(x’, y’) = (2, -3)
Let (h, k) be the shifting 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) x² + y² + 2x – 4y.+ 1 = 0
ii) 2x² + y² – 4x + 4y = 0
Solution:
i) The given equation is
x² + y² + 2x – 4y + 1 = 0
Origin is shifted to (-1, 2)
h = -1, k = 2
Equation of transformations are
x = x’ + h, y = y’ + k
i.e., 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 + 4 + 4y’ + 2x’ – 2 -4y’ – 8 + 1 = 0
(x’)² + (y’)² -4 = 0
The transformed equation is x² + y² – 4 = 0

ii) Old equation is
2x² + y² – 4x + 4y = 0
New equation is 2(x’ – 1)² + (y’ + 2)² —4(x’ – 1) + 4(y’ + 2) = 0
2[(x’)² + 1 – 2x’] + (y’)² + 4 + 4y’ – 4x’ + 4 + 4y’ + 8 = 0
2(x’)² + 2 – 4x’ + (y’)² + 4 + 4y’ – 4x’ + 4 + 4y’ + 8 = 0
2(x’)² + (y’)² – 8x’ + 8y’ + 18 = 0
The transformed equation is
2x² + y² – 8x + 8y+18 = 0

Question 5.
The point to which the origin is shifted and the transformed equation are given below. Find the original equation.
i) (3,-4);x² + y² = 4
ii) (-1, 2); x² + 2y² + 16 = 0
Solution:
i) Given shifting origin = (3, – 4) = (h, k)
x’ = x – h,
= x – 3

y’ = y – k
= y + 4

The original equation of (x’)² + (y’)² = 4 is
(x – 3)² + (y + 4)2 = 4
x² – 6x + 9 + y² + 8y + 16 = 4
x² + y² – 6x + 8y + 21 =0

ii) Given shifting origin = (h, k) = (-1,2)
x’ = x – h,
= x + 1

y’ = y – k
y = y – 2

The original equation of
x’² + 2y’² + 16 = 0 is
(x + 1)² + 2(y – 2)² + 16 = 0
x² + 2x + 1 + 2y² – 8y + 8 + 16 = 0
x² + 2y² + 2x – 8y + 25 = 0 Question 6.
Find the point to which the origin is to be shifted so as to remove the first degree terms from the equation.
4x² + 9y² – 8x + 36y + 4 = 0
Solution:
The given equation is
4x² + 9y² – 8x + 36y + 4 = 0
a = 4 g = -4
b = 9 f = 18 Origin should be shifted to (1, -2)

Question 7.
When the axes are rotated through an angle 30°, find the new coordinates of the following points,
i) (0, 5) ii) (-2, 4) hi) (0, 0)
Solution:
i) Given 0 = 30°
Old co-ordinates are (0, 5)
i.e., x = 0, y = 5
x‘ = x. cos θ + y. sin θ
= 0. cos 30° + 5. sin 30° = $$\frac{5}{2}$$
= – x sin θ + y cos θ
= – 0. sin 30° + 5 cos 30° =
New co-ordinates are $$\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$$

ii) Old co-ordinates are (-2, 4)
x= -2, y = 4
x’ = x cos θ + y sin θ
= (-2). cos 30°+ 4. sin 30°
= -2. $$\frac{\sqrt{3}}{2}$$ + 4. $$\frac{1}{2}$$ = – √3 +2
y’ = -x sin θ + y cos θ
= – (-2) sin 30° + 4 cos 30°
= 2 . $$\frac{1}{2}$$ + 4. $$\frac{\sqrt{3}}{2}$$
= 1 + 2 √3
New co-ordinates are (- √3 + 2, 1 + 2√3)

iii) Given (x, y) = (0,0) and 0 = 30°
x = (0, y) ⇒ x = x’. cos 30° – y’ sin 30°
= 0. $$\frac{\sqrt{3}}{2}$$ – 0. $$\frac{1}{2}$$ =0
y = x’. sin 30° + y’.cos 30°
= 0.$$\frac{1}{2}$$ + 0.$$\frac{\sqrt{3}}{2}$$ = 0
New co-ordinates of the point are (0, 0)

Question 8.
When the axes are rotated through an angle 60°, the new co-ordinates of three points are the following
i) (3, 4) ii) (-7, 2) iii) (2, 0) Find their original coordinates.
Solution:
i) Given 0 = 60°
New co-ordinates are (3, 4)
x’ = 3, y’ = 4
x = x’ cos θ – y’ sin θ
= 3. cos 60° – 4. sin 60°
$$=3 \cdot \frac{1}{2}-\frac{4 \cdot \sqrt{3}}{2}=\frac{3-4 \sqrt{3}}{2}$$
y = x’ sin θ + y’ cos θ
= 3 sin 60° + 4. cos 60° ii) New co-ordinates are (-7, 2)
x’= -7, y’ = 2
x = x’ cos θ – y’ sin θ
= (-7) cos 60° – 2. sin 60°
$$=-7 \cdot \frac{1}{2}-2 \cdot \frac{\sqrt{3}}{2}=\frac{-7-2 \sqrt{3}}{2}$$
y = x’ sin θ + y’. cos θ
= – 7. sin 60° + 2. cos 60° iii) New co-ordinates are (2, 0)
x’ = 2, y’ = 0
x = x’ cos θ – y’ sin θ
= 2. cos 60° – 0. sin 60°
= 2.$$\frac{1}{2}$$ – 0.$$\frac{\sqrt{3}}{2}$$ =1 – 0 = 1
y = x’ sin θ + y’ cos θ
= 2. sin 60° + 0. cos 60°
= 2.$$\frac{\sqrt{3}}{2}$$ + 0.$$\frac{1}{2}$$ = √3
Co-ordinates of R are (1, √3) Question 9.
Find the angle through which the axes are to be rotated so as to remove the xy term in the equation.
x² + 4xy + y² – 2x + 2y – 6 = 0.
Solution:
Compare the equation
x² + 4xy + y² – 2x + 2y – 6 = 0
with ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 1, h = 2, b = 1, g = -1, 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 x² + 3xy – 2y² + 17x – 7y – 11= 0. Find the original equation of the curve.
Solution:
Equations of transformation are
x = x’ + h, y = y’ + k
x’ = x – h = x – 2, y’ = y – 3
Transformed equation is
x² + 3xy – 2y² + 17x – 7y – 11 = 0
Original equation is
(x – 2)² + 3(x – 2) (y – 3) – 2(y – 3)² + 17(x – 2) – 7(y – 3) – 11 = 0
⇒ x² – 4x + 4 + 3xy – 9x – 6y + 18 – 2y² + 12y – 18 + 17x – 34 – 7y + 21 -11 = 0
⇒ x² + 3xy – 2y² + 4x – y – 20 = 0
This is the required original equation.

Question 2.
When the axes are rotated through an angle 45°, the transformed equation of acurveis 17x² – 16xy + 17y² = 225. Find the original equation of the curve.
Solution:
Angle of rotation = θ = 45
x’ = x cos θ + y sin θ = x cos 45 + y sin 45 = $$\frac{x+y}{\sqrt{2}}$$
y’ = – x sin θ + y cos θ = – x sin 45 + y cos 45 = $$\frac{-x+y}{\sqrt{2}}$$

The original equation of
17x’²- 16x’y’ + 17y’² = 225 is ⇒ 17x² + 17y² + 34xy – 16y² + 16x² + 17x² + 17y² – 34xy = 450
⇒ 50x² + 18y² = 450
∴ x² + y² = 9 is the original equation.

Question 3.
When the axes are rotated through an angle a, find the transformed equation of x cos a + y sin a = p. iMM&mmn
Solution:
The given equation is x cos α + y sin α = p
∵ The 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’ (cos² α + sin² α) = p
⇒ x’ = p
The equation transformed to x = p

Question 4.
When the axes are rotated through an angle n/6. Find the transformed equation of x² + 2 √3xy – y² = 2a².
Solution: ⇒ 3X² – 2√3 XY + Y² + 2√3|√3X² +2XY – √3Y²|- (x² +3Y² +2√3XY) =8a²
⇒ 3X² -2V3XY +Y² +6X² + W3XY – 6Y² – X² – 3Y² – 2√3 XY = 8a²
⇒ 8X² – 8Y² = 8a² ⇒ X² – Y² = a²
The transformed equation is x² – y² = a² Question 5.
When the axes are rotated through an angle $$\frac{\pi}{4}$$, find the transformed equation of 3x² + 10xy + 3y² = 9.
Solution:
Given equation is
3x² + 10xy + 3y² – 9 = 0 ………….. (1)
Angle of rotation of axes = θ = $$\frac{\pi}{4}$$
Let (X, Y) be the new co-ordinates of (x, y)
x = X cos θ – Y sin θ Transformed equation of (1) is ⇒ 3X² – 6XY + 3Y² + 10X² – 10Y² + 3X² + 6XY + 3Y² – 18 = 0
∴ 16X² – 4Y² -18 = 0
∴ 8X² – 2Y² = 9
∴ 8X² – 2Y² = 9
The transformed equation is 8x² – 2y² = 9