Use these Inter 1st Year Maths 1B Formulas PDF Chapter 2 Transformation of Axes to solve questions creatively.

## Intermediate 1st Year Maths 1B Transformation of Axes Formulas

→ x = x’ + h ⇒ x’ = x – h

→ y = y’ + k ⇒ y’ = y – k

→ To make the first degree term absent, origin should be shifted to \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If the equation is ax^{2} + by^{2} + 2gx + 2fy + c = 0, origin should be shifted to be \(\left(-\frac{g}{a},-\frac{f}{b}\right)\)

x’ | y’ | |

x | cos θ | – sin θ |

y | sin θ | cos θ |

x = x’ cos θ – y’ sin θ

x’ = x cos θ + y sin θ

y = x’ sin θ + y’ cos θ

y’ = – x sin θ + y cos θ

→ To make the xy term absent, axes should be rotated through an angle θ given by

tan 2θ = \(\frac{2 h}{a-b}\)

**Rotation Of Axes (Change Of Direction):**

I. Definition: If thc axes are rotated through an angle in the same plane by keeping the origin constant, then the transformation is called Rotation of axes.

2. Theorem: To find the co-ordinates of a point (x,y)are transformed to (X, Y)when the axes are rotated through an angle ‘6’ about the origin in the same plane.

Proof: Let x’Ox, yOY’ are the original axes

Let P(x,y)be the co-ordinates of the point in the above axes.

After rotating the axes through an angle ‘θ’, then the co-ordinates of P be (X, Y) w.r.t the new axes X’OX and YOY’ as in figure.

Since θ is the angle of rotation, then ∠xOA = ∠yOY = θ as in the figure.

Since L, M is projections of P on Ox and OX respectively. We can see that ∠LPM = ∠xOA = θ

Let N be the projection to PL from M

Now x = OL = OQ – LQ = OQ – NM

= OM cos θ – PM sin θ

= A cos θ – Y sin θ

y = PL = PN + NL = PN + MQ

PM cos θ + OM sin θ = Ycos θ + X sin θ

x = X cos θ – Y sin θ and

y = Y cos θ + X sin θ ………….(1)

Solving the above equations to get X and Y, then X = x cos θ + y sin θ and Y = -x sin θ + y cos θ ………… (2)

From (1) and (2) we can tabulate

x’ | y’ | |

x | cos θ | – sin θ |

y | sin θ | cos θ |

Note:

- If the axes are turned through an angle ‘ θ ’, then the equation of a curve f (x, y) = 0 is transformed to f (X cos θ – Y sin θ, X sin θ + Y cos θ) = 0
- If f (X, Y) = 0is the transformed equation of a curve when the axes are rotated through an angle ‘ θ ’ then the original equation of the curve is f (x cos θ + y sin θ, – x sin θ + y cos θ) = 0

**Theorem:**

To find the angle of rotation of the axes to eliminate xy term in the equation ax^{2} + 2hxy + by^{2} + 2 gx + 2 fy + c = 0

Proof: given equation is ax^{2}+ 2hxy + by^{2} + 2 gx + 2 fy + c = 0

Since the axes are rotated through an angle θ, then x = X cos θ – Y sin θ, y = X sin θ + Y cos θ

Now the transformed equation is

a (X cos θ – Y sinθ)^{2} + 2h (X cos θ – Y sin θ)(X sin θ + Y cos θ) + b (X sin θ + Y cos θ)^{2} + 2g (X cos θ – Y sin θ) + 2 f (X sin θ + Y cos θ) + c = 0

⇒ X^{2}cos^{2}θ + Y^{2}sin^{2}θ – 2XYcosθsinθ) + 2h [X^{2}cos θ + XY(cos^{2}θ – sin^{2}θ) – Y sin^{2}θ cos^{2}θ] + b[X^{2} sin^{2}θ + Y^{2} cos^{2}θ + 2XY cos θ sin θ) + 2g (X cos θ – Y sin θ) + 2 f (X sin θ + Y cos θ) + c = 0

Since XY term is to be eliminated, coefficient of XY = 0.

⇒ 2 (b – a) cos θ sin θ + 2h (cos2 θ – sin2 θ) = 0

⇒ h cos 2θ + (b – a) sin 2θ = 0

⇒ 2hcos2θ = (a -b)sin 2θ

⇒ Angle of rotation (θ) = \(\frac{1}{2}\) Tan^{-1}\(\left(\frac{2 h}{a-b}\right)\)

Note: The angle of rotation of the axes to eliminate xy term in

ax^{2} + 2hxy + ay^{2} + 2 gx + 2 fy + c = 0 is \(\frac{\pi}{4}\)

**Translation of Axes:**

1. Definition: Without changing the direction of the axes, the transformation in which the origin is shifted to another position or point is called translation of axes.

2. Theorem: To find the co-ordinates of a point (x, y) are translated by shifting the origin to a point (x, y)

Proof :

Let xox^{1} xox^{1}, yoy^{1} be the original axes and A(x_{1}, y_{1}) be a point to which the origin is shifted

Let AX, AY be the new axes which are parallel to the original axes as in figure.

Let P be any point in the plane whose coordinates w.r.t old system are (x, y)

And w.r.t new axes are (X, Y) .

From figure, A = (x_{1}, y_{1}) then AL = y_{1}, OL = x_{1},

P(x, y )then x = ON = OL + LN = OL + AM = x_{1} + X = X + x_{1}

Hence x = X + x_{1}.

y = PN = PM +MN = X+ AL = X + y_{1}

therefore, y = Y + y_{1}

hence x = X + x_{1} y = Y + y_{1}

3. Theorem: To find the point to which the origin is to be shifted by translation of axes to eliminate x, y terms(first degree terms) in the equation ax^{2} + 2xhy + by^{2} + 2gx + 2fy + c = 0 (2 = ab)

Proof : given equation is ax2 + 2 xhy + by2 + 2gx + 2fy + c = 0

Let (x1, y1) be a point to which the origin is shifted by translation Let (X ,Y) be the new co-ordinates of the point (x, y) .

the equations of the transformation are x = X + x_{1}, y = Y + y_{1}

Now the transformed equation is

a (X + x_{1})^{2} + 2h (X + x_{1})(Y + y_{1})+ b (Y + y_{1})^{2} + 2g (X + x_{1}) + 2f [Y + y_{1}] + c = 0

⇒ a(X^{2} + 2x_{1}X + x_{1}^{2}) + 2h(XY + x_{1}Y + y_{1}X + x_{1}y_{1}) + b(Y^{2} + 2y_{1}Y + y_{1}^{2}) + 2g(X + x_{1}) + 2f (Y + y_{1}) + c = 0

⇒ aX^{2} + 2hXY + bY^{2} + 2X [ax_{1} + hy_{1} + g) + 2Y(hx_{1} + by_{1} + f ) + (ax_{1}^{2} + 2x_{1} + by_{1}^{2} + 2gx_{1} + 2fy_{1} + c) = 0

Since x, y terms (the first degree terms) are to be eliminated

ax_{1} + hy_{1} + g = 0

hx_{1} + by_{1} + f = 0

Solving these two equations by the method of cross pollination

\(\frac{x_{1}}{h f-b g}=\frac{y_{1}}{g h-a f}=\frac{1}{a b-h^{2}}\) ⇒ x_{1} = \(\frac{h f-b g}{a b-h^{2}}\) y_{1} = \(\frac{g h-a f}{a b-h^{2}}\)

New Origin is ⇒ (x_{1}, y_{1}) = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Note:

(i) The point to which the origin has to be shifted to eliminate x, y terms by translation in the equation ax2 + by2 + 2gx + 2 fy + c = 0is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

If b = a, then the new origin is \(\left(\frac{-g}{a}, \frac{-f}{b}\right)\)

(ii) The point to which the origin has to be shifted to eliminate x, y terms by translation of axes in the equation a(x + x_{1})^{2} + b(y + y_{1})^{2} = c is (-x_{1}, – y_{1})

(iii) The point to which the origin has to be shifted to eliminate x, y terms by translation in the equation 2hxy + 2gx + 2 fy + c = 0 is \(\left(\frac{-f}{h}, \frac{-g}{h}\right)\)

l h ’ h I

Theorem : To find the condition that the equation ax^{2} + 2hxy + by^{2} + 2 gx + 2 fy + c = 0 to be in the form aX^{2} + 2hXY + bY^{2} = 0 when the axes are translated. (h^{2} ≠ ab)

Proof : From theorem 3, we get

ax_{1} + hy_{1} + g = 0 ………..(1)

hx_{1} + by_{1} + f = 0 …………(2)

Solving (1) and (2),

(x_{1}, y_{1}) = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g f-a f}{a b-h^{2}}\right)\)

Since the equation is to be in the form of aX^{2} + 2hXY + bY^{2} = 0 ,then for this we should have axj^{2} + 2hx_{1}y_{1} + by^{2} + 2 gx_{1} + 2 fy_{1} + c = 0

⇒ (ax_{1} + hyx + g) x_{1} + (x_{1} + byx + f) yx + (gx + fy_{1} + c ) = 0

⇒ (0).x + (0), y + (gx_{1} + fyx + c) = 0 from(1) and (2) ⇒ gx_{1} + fyx + c = 0 …………..(3)

Substituting x_{1}, y_{1 }in (3), we get

g\(\left(\frac{h f-b g}{a b-h^{2}}\right)\) + f\(\left(\frac{g h-a f}{a b-h^{2}}\right)\) + c = 0

⇒ fgh – bg^{2} + fgh – af^{2} + abc + -ch^{2} = 0

⇒ abc + 2 fgh – af^{2} – bg^{2} – ch^{2} = 0