Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Transformation of Axes Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1B Transformation of Axes Important Questions

Question 1.

When the origin is shifted to (2, 3) by the translation of axes, the coordinates of a point p are changed as (4, -3). Find the coordinates of P in the original system.

Solution:

(h, k) = (2, 3) ⇒ h = 2, k = 3

(x’, y’) = (4, 3) ⇒ x = 4, y = -3

x = x’ + h = 4 + 2 = 6, y = y’ + k = -3 + 3 = 0

original Co-ordinates are (6, 0)

Question 2.

Find the point to which the origin is to be shifted by the translation of axes so as to remove the first-degree terms from the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, where h^{2} ≠ ab.

Solution:

Let the origin be shifted to (α, β) by the translation of axes.

Then x = x’ + α,

y = y’ + β

On substituting these in the given equation, we get

a(x’ + α)^{2} + 2h(x’ + α) (y’ + β) + b(y’ + β)^{2} + 2g(x’ + α) + 2f(y’ + β) + c = 0

Which gives

ax’^{2} + 2hx’y’ + by^{2} + 2x'(α + β + g) + 2y'(hα + bβ + f) + aα^{2} + 2hαβ + bβ^{2} + 2gα + 2fβ + c = 0 ……….. (1)

If equation (1) has to be free from the first degree terms, then we have

aα + hβ + g = 0 and hα + bβ +1 = 0

Solving these equations for a and 13, we get

α =\(\frac{h f-b g}{a b-h^{2}}\) , β = \(\frac{g h-a f}{a b-h^{2}}\)

Therefore, the origin is to be shifted to

\(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Question 3.

Find the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation

ax^{2} + by^{2} + 2gx + 2fy + c = 0, where a ≠ 0, b ≠ 0.

Solution:

Here the given equation does not contain xy term. Hence writing h = 0 in the result of

Problem 12, the required point is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

Question 4.

If the point P changes to (4, -3) when the axes are rotated through an angle of 135°, find the coordinates of P with respect to the original system.

Solution:

Here (x’, y’) = (4, -3); θ = 135°

Let (x, y) be the coordinates of P, then

x = x’ cos θ – y’ sin θ

= 4cos 135° – (-3) sin 135°

Therefore, the coordinates of P with respect to the original system are (\(\frac{-1}{\sqrt{2}}\), \(\frac{7}{\sqrt{2}}\))

Question 5.

Show that the axes are to be rotated through an angle of \(\frac{1}{2}\) Tan^{-1}(\(\frac{2 h}{a-b}\)) so as to remove the xy term from the equation ax^{2} + 2hxy + by^{2} = 0, if a ≠ b and through the angle \(\frac{\pi}{4}\), if a = b. [Mar 13, 06]

Solution:

If the axes are rotated through an angle ‘θ’,

then

x = x’ cos θ – y’ sin θ

y = x’ sin θ + y’ cos θ

Therefore the given equation transforms as

a(x’ cos θ – y’ sin θ)^{2} + 2h(x’ cosθ – y’ sinθ) (x’ sinθ + y’ cosθ) + b(x’ sin θ + y’ cos θ)^{2} = o

To remove x’y’ term from the equation, the coefficient of x’y’ term must be zero.

So, (b – a) sin θ cos θ + h(cos^{2}θ – sin^{2}θ) = 0

i.e., h cos 2θ = \(\frac{a-b}{2}\) sin 2θ

i.e., tan 2θ = \(\frac{2 h}{a-b}\), if a ≠ b and h cos 2θ = 0, if a = b

Therefore θ = \(\frac{1}{2}\) Tan^{-1}(\(\frac{2 h}{a-b}\)), if a ≠ b and θ = \(\frac{\pi}{4}\), If a = b

Question 6.

When the origin is shifted to (-2, -3) and the axes are rotated through an angle 45° find the transformed equation of 2x^{2} + 4xy – 5y^{2} + 20x – 14 = 0.

Solution:

Here (h, k) = (-2, -3), h = -2, k = -3

θ = 45°

Let (x’, y’) be the new co-ordinates of any point (x, y) is the plane after transformation

x = x’ cos θ – y’ sin θ + h = -2 + x’ cos 45° – y’ sin 45°.

= -2 + \(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\)

y = x’ si nθ + y’ cos θ + k = x’ sin 45° + y’ cos

45° – 3 = -3 + \(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\)

The transformed equation is

⇒ (x’ – y’)^{2} + 8 – 4\(\sqrt{2}\) (x’ – y’) + 2(x’^{2} – y’^{2}) – 6\(\sqrt{2}\) (x’ – y’) – 4\(\sqrt{2}\) (x’ + y’) + 24 = 0

⇒ –\(\frac{5}{2}\) (x’ – y’)^{2} – 45 + 15\(\sqrt{2}\) (x’ + y’) + 10\(\sqrt{2}\) (x’ – y’) – 40 – 11\(\sqrt{2}\) (x’ + y’) + 66 – 14 = 0

⇒ x’^{2} + y’^{2} – 2x’y’ + 2x’^{2} – 2y’^{2} – \(\frac{5}{2}\) (x’^{2} + y’^{2} + 2x’y’) – 1 = 0

\(\frac{1}{2}\) x’^{2} – \(\frac{7}{2}\) y’^{2} – 7x’y’ – 1 = 0

i.e., x’^{2} – 7y’^{2} – 14x’y’ – 2 = 0

The transformed equation is (dropping dashes)

x^{2} – 7y^{2} – 14xy – 2 = 0

Question 7.

When the origin is shifted to (-2, 3) by translation of axes, let us find the co-ordinates of (1, 2) with respect to new axes.

Solution:

Here (h, k) = (-2, 3)

Let (x, y) = (1, 2) be shifted to (x’ y’) by the translation of axes.

Then (x’, y’) = (x – h, y – k)

= (1 – (-2), 2 – 3) = (3, -1)

Question 8.

When the origin is shifted to (3, 4) by the translation of axes, let us find the transformed equation of 2x^{2} + 4xy + 5y^{2} = 0.

Solution:

Here (h, k) = (3, 4)

on substituting x = x’ + 3 and y = y’ + 4 in the given equation.

as per note the equation f(x, y) = 0 of the curve is transformed as f(x’ + h, y’ + k) = 0]

we get

2(x’ + 3) + 4(x’ + 3) (y’ + 4) + 5(y’ + 4)^{2} = o

Simplifying this equation, we get

2x’^{2} + 4x’y’ + 5y’^{2} + 28x’ + 52y’ + 146 = 0

This equation can be written (dropping dashes) as

2x^{2} + 4xy + 5y^{2} + 28x + 52y + 146 = 0