# Translation and Rotation of Axes Worksheet - Page 2

Translation and Rotation of Axes Worksheet
• Page 2
11.
Identitfy the type of conic represented by the equation 6$x$2 - 4$x$$y$ + 7$y$2 - 40$x$ + 60$y$ - 47 = 0.
 a. a circle b. an ellipse c. a parabola d. a hyperbola

#### Solution:

6x2 - 4xy + 7y2 - 40x + 60y - 47 = 0

A = 6, B = - 4, and C = 7
[Compare with Ax² + Bxy + Cy² + Dx + Ey + F = 0.]

Discriminant is B2 - 4AC = (- 4)2 - 4(6)(7)

= - 152 < 0

So, the equation represents an ellipse.
[B2 - 4AC < 0.]

12.
Identify the type of conic represented by the equation 6$x$2 - 7$y$2 - 2$y$ - 34 = 0.
 a. a parabola b. an ellipse c. a circle d. a hyperbola

#### Solution:

6x2Ã‚Â - 7y2Ã‚Â - 2y - 34 = 0

A = 6, B = 0, and C = - 7
[Compare with Ax² + Bxy + Cy² + Dx + Ey + F = 0.]

Discriminant is B2 - 4AC = (0)2 - 4(6)(- 7)

= 168 > 0

So, the given equation represents a hyperbola.
[B2 - 4AC > 0.]

13.
Choose the equation in standard form for the given conic. [Given $a$ = 4, $b$ = 3.]

 a. $x$2 + $y$2Ã‚Â = 144 b. 9$x$2 + 16$y$2 = 144 c. 16$x$2Ã‚Â + 9$y$2Ã‚Â = 1 d. $x$2Ã‚Â + 16$y$2 = 1

#### Solution:

The given conic is an ellipse having semi major axis a = 4 and semi minor axis b = 3
[From the figure.]

x2a2 + y2b2 = 1
[Write the standard form of an ellipse.]

So, the equation of the conic shown is x216 + y29 = 1 or 9 x2 + 16 y2 = 144.
[Substitute the values of a, b in the standard form and simplify.]

14.
Choose the co-ordinates of P(5, - 2) in the rotated system, if the axes are rotated through an angle $\frac{\pi }{3}$.
 a. Does not exist b. $x$′ = c. $x$′ = , $y$′ = - d. $x$′ = , $y$′ = -

#### Solution:

If the axes is rotated through an angle α, then the co-ordinates of P(x, y) in the new system are x′ = xcos α + ysin α and y′ = - xsin α + ycos α.

x′ = 5cos (π3) + (- 2)sin (π3) and y′ = - 5sin (π3) + (- 2)cos (π3)
[Substitute α = π3, P(x, y) = (5, - 2).]

x′ = 5 - 232 and y′ = - (53 + 2)2

So, the co-ordinates of P in the rotated system are (5 - 232, - (53 + 2)2).

15.
Identify the transformed equation of $x$² + 2$\sqrt{3}$$x$$y$ - $y$² = 2$a$², when the axes are rotated through an angle $\frac{\pi }{6}$.
 a. $x$′² + $y$′² = $a$² b. $x$′² + $y$′² = - $a$² c. $x$′² - $y$′² = - $a$² d. $x$′² - $y$′² = $a$²

#### Solution:

When the axes are rotated through an angle α, then the coordinates of a point P(x, y) with respect to the original axes are related to its coordinates (x', y') with respect to the rotated axes as x = x′ cos α - y′ sin α and y = x′ sin α + y′ cos α.

x = x′ cos (π6) - y′ sin (π6) = 3x - y2
[Substitute α = π6.]

y = x′ sin (π6) + y′ cos (π6) = x +3y2
[Substitute α = π6.]

x² + 23xy - y² = 2a²
[Original equation.]

(3x-y2)² + 23(3x-y2)(x+3y2) - (x+3y2)² = 2a²

(3x-y)2+23(3x-y)(x+3y) - (x+3y)2 = 8a²

x2 - y2 = a²
[Simplify.]

When the axes are rotated through an angle π6, the transformed equation of x² + 23xy - y² = 2a² becomes x′² - y′² = a².

16.
Find the approximate angle of rotation needed to eliminate the cross product (or) $x$' $y$' term in the transformed equation of 4$x$2 + 2$x$$y$ + 2$y$2 - 47 = 0.
 a. $\alpha$ = cot-1 (1 ) b. $\alpha$ = $\frac{1}{2}$ tan-1 (1 ) c. $\alpha$ = $\frac{1}{2}$ cot-1 (1 ) d. $\alpha$ = tan-1 (1 )

#### Solution:

4x2 + 2xy + 2y2 - 47 = 0

A = 4, B = 2, and C = 2
[Compare with Ax² + Bxy + Cy² + Dx + Ey + F = 0.]

Let α be the angle of rotation needed to eliminate the x'y' - term in the transformed equation of the original equation, then cot 2α = A - C / B

2α = cot-1 (A - C / B)
[Solve for 2α.]

2α = cot-1 (4 - 22)
[Substitute the values of A, B, and C.]

α = 1 / 2 cot-1 (1 )
[Solve for α by using a calculator.]

17.
Choose the conic represented by the equation 8$y$2 + 48$y$ - 8$x$2 + 32$x$ = 24.
 a. an ellipse b. a parabola c. a circle d. a hyperbola

#### Solution:

8y2 + 48y - 8x2 + 32x = 24
[Original equation.]

8(y + 3)2 - 8(x - 2)2 = 64
[Group the terms and complete the squares.]

(y + 3)28 - (x - 2)28 = 1
[Divide on both sides by 64.]

The translated equation of the original equation becomes
(y)28 - (x)28 = 1 which represents a hyperbola.
[Translate the axes using h = 2 and k = - 3 and use x′ = x - 2, y′ = y + 3.]

So, the conic represented by the original equation is a hyperbola.

18.
Write an equation in standard form for the conic shown.

 a. 9$x$2Ã‚Â - 16$y$2 = - 144 b. 9$x$2Ã‚Â - 16$y$2 = 144 c. 9$x$2Ã‚Â + 16$y$2 = 144 d. 16$x$2Ã‚Â - 9$y$2 = 144

#### Solution:

The given conic section represents a hyperbola x2a2 - y2b2 = 1.

x232 - y242 = 1
[From the figure, a = 3, b = 4.]

16x2 - 9y2Ã‚Â = 144

So, the equation of the conic shown is x2 - 9y2Ã‚Â = 144.

19.
Which of the following equations represents the standard form of the conic 16$x$2 - $y$2 - 32$x$ - 6$y$ - 57 = 0? Translate the axes so that the origin is at the center.
 a. $\frac{{\left(x\prime \right)}^{2}}{{2}^{2}}$ - $\frac{{\left(y\prime \right)}^{2}}{{8}^{2}}$ = 1 b. $\frac{{\left(x\prime \right)}^{2}}{{8}^{2}}$ - $\frac{{\left(y\prime \right)}^{2}}{{2}^{2}}$ = 1 c. $\frac{{\left(x\prime \right)}^{2}}{{8}^{2}}$ + $\frac{{\left(y\prime \right)}^{2}}{{2}^{2}}$ = 1 d. None of the above

#### Solution:

16x2 - y2 - 32x - 6y - 57 = 0
[Original equation of the conic.]

16(x2 - 2x + 1) - (y2 + 6y + 9) - 64 = 0
[Group the terms.]

16(x - 1)2 - (y + 3)2 = 64
[Complete the squares.]

(x - 1)24 - (y + 3)264 = 1
[Divide both sides by 64.]

(x - 1)222 - (y + 3)282 = 1

The transformed equation of the original equation is (x)222 - (y)282 = 1, which is in the standard form.
[Translate the conic using h = 1 and k = -3 and use x′ = x - 1, y′ = y + 3.]

20.
Which of the following is true about the conic A$x$2 + C$y$2 + D$x$ + E$y$ + F = 0?
 a. The focal axis of the given equation is not aligned with one of the co-ordinate axes. b. It represents an ellipse. c. It represents a line. d. The focal axis of the given equation is aligned with one of the co-ordinate axes.

#### Solution:

Ax2Ã‚Â + Cy2 + Dx + Ey + F = 0
[Original equation of the conic.]

It is clearly known from the equation that the cross product term (xy-term) is missing.

Hence, the focal axis of the given conic is aligned with one of the co-ordinate axes.