﻿ Rotation of Conics Worksheet | Problems & Solutions

# Rotation of Conics Worksheet

Rotation of Conics Worksheet
• Page 1
1.
Choose the coordinates of a point (- 4, - 4) when the axes are turned through an angle 90°.
 a. (4, 4) b. (4, - 4) c. (- 4, 4) d. (- 4, - 4)

#### Solution:

When the axes are turned through an angle θ, then the new coordinates in terms of original coordinates are x′ = x cos θ + y sin θ and y′ = -x sin θ + y cos θ.
[Formula.]

x′ = (- 4) cos 90° + (- 4) sin 90°
[Substitute the values.]

= (- 4)(0) + (-4)(1) = - 4

y′ = -(-4) sin 90° + (-4) cos 90°
[Substitute the values.]

= (4)(1) + (-4)(0) = 4

The new coordinates are (- 4, 4).

2.
Choose the original coordinates of a point P in terms of new coordinates, when the axes are rotated through an angle θ without changing the origin, where ($x$, $y$) and ($x$′, $y$′) be the original and new coordinates.
 a. $x$ = $x$′ cos $\theta$ and $y$ = $y$′ sin $\theta$ b. $x$′ = $x$ cos $\theta$ and $y$′ = $y$ sin $\theta$ c. $x$ = $x$′ cos $\theta$ - $y$′ sin $\theta$ and $y$ = $x$′ sin $\theta$ + $y$′ cos $\theta$ d. None of the above

#### Solution:

When the axes are rotated through an angle θ, then the original coordinates in terms of new coordinates will be x = x′ cos θ - y′ sin θ and y = x′ sin θ + y′ cos θ.

3.
The transformed equation of a conic after the axis are rotated through some angle is given by $x$2 + 48$\sqrt{3}$$\mathrm{xy}$ + 2$y$2 + (5 + 2$\sqrt{3}$)$x$ + (5$\sqrt{3}$ - 2)$y$ + 48 = 0. The conic is ________.
 a. ellipse b. circle c. hyperbola d. parabola

#### Solution:

x2 + 483xy + 2y2 + (5 + 23)x + (53- 2)y + 48 = 0 is the given transformed equation of a conic.

Comparing the given equation with Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we get A = 1, B = 483, and C = 5.

Discriminant = B2 - 4AC = = (483)2 - 4(1) (2) = 6904

Since B2 - 4AC > 0, the given conic equation represents a hyperbola.

4.
Choose the new coordinates of a point P in terms of original coordinates, when the axes are rotated through an angle θ without changing the origin, where ($x$, $y$) and ($x$′, $y$′) be the original and new coordinates.
 a. $x$′ = $x$ cos $\theta$ + $y$ sin $\theta$ and $y$′ = -$x$ sin $\theta$ + $y$ cos $\theta$ b. $x$ = $x$′ cos $\theta$ and $y$ = $y$′ sin $\theta$ c. $x$′ = $x$ cos $\theta$ and $y$′ = $y$ sin $\theta$ d. None of the above

#### Solution:

When the axes are rotated through an angle θ then the new coordinates of a point P in terms of original coordinates are x′ = x cos θ + y sin θ and y′ = -x sin θ + y cos θ.

5.
Choose the angle to which the axes are to be rotated so that the $\mathrm{x\text{'} y\text{'}}$ term in the translated equation of A$x$2 + B$\mathrm{xy}$ + C$y$2 + D$x$ + E$y$ + F = 0 will not be present.
 a. $\theta$ = $\frac{1}{2}$${\mathrm{Cot}}^{-1}\left[\frac{A-C}{B}\right]$ b. $\theta$ = ${\mathrm{cot}}^{-1}\left[\frac{A-C}{B}\right]$ c. $\theta$ = $\frac{1}{2}$${\mathrm{cot}}^{-1}\left[\frac{A-B}{C}\right]$ d. $\theta$ = $\frac{1}{2}$${\mathrm{cot}}^{-1}\left[\frac{A}{B}\right]$

#### Solution:

The x' y' - term in the transformed equation will be removed when the axes are turned through an angle θ = 1 / 2 Cot-1[A-C / B].

6.
Choose the coordinates of the point P (7, 0) in original system, when the original system is rotated through an angle $\frac{\pi }{4}$.
 a. ($\frac{7}{\sqrt{2}}$, 0) b. (0, $\frac{7}{\sqrt{2}}$) c. ($\frac{7}{\sqrt{2}}$, $\frac{7}{\sqrt{2}}$) d. ($\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$)

#### Solution:

When the axes are rotated through an angle θ, then the original coordinates in terms of new coordinates will be x = x′ cos θ - y′ sin θ and y = x′ sin θ + y′ cos θ.

x = x′ cos θ - y′ sin θ

= (7) cos (π4) - (0) sin (π4)
[Substitute the values.]

= 72 - 0 = 72

y = x′ sin θ + y′ cos θ

= (7) sin (π4) + (0) cos(π4)
[Substitute the values.]

= 72 + 0 = 72

Hence, the coordinates of the point P in the original system are (72, 72).

7.
Choose the coordinates of the point P (0, 10) in the original system, when the original system is rotated through an angle $\frac{\pi }{3}$.
 a. [ 5 $\sqrt{3}$, 5 ] b. [ $\sqrt{3}$, 5 ] c. [- 5 $\sqrt{3}$, 5 ] d. [ 5 $\sqrt{3}$, - 5 ]

#### Solution:

When the axes are rotated through an angle θ, then the original coordinates in terms of new coordinates will be x = x′ cos θ - y′ sin θ and y = x′ sin θ + y′ cos θ.
[Formula.]

x = x′ cos θ - y′ sin θ

= (0) cos (π3) - (10) sin (π3)
[Substitute the values.]

= - 5 3

y = x′ sin θ + y′ cos θ

= (0) sin (π3) + (10) cos (π3)
[Substitute the values.]

= 5

Hence, the coordinates of point P in original system is [- 5 3, 5 ].

8.
Choose the conic, which is represented by the equation $x$2 - $\mathrm{xy}$ + $y$2 = 2 when it is rotated through a suitable angle.
 a. Ellipse b. Hyperbola c. Parabola d. None of the above

#### Solution:

x2 - xy + y2 = 2

Comparing the given equation with Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, we get A = 1, B = -1, C = 1, D = 0, E = 0, and F = -2.

So, θ = 1 / 2 Cot-1[A-C / B] θ = 1 / 2 Cot-1[1-1-1]

= 12 Cot-1(0) = π4

The second degree equation obtained when we apply rotation to Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is A′x2 + B′xy + C′y2 + D′x + E′y + F′ = 0 where the values of coefficients are calculated as follows.
[Formula.]

A′ = (1)( 1 / 2) + (-1)( 1 / 2) + (1)( 1 / 2) = 1 / 2
[Substitute in A′ = A cos² θ + B cos θ sin θ + C sin² θ.]

B′ = (-1) (0) + (0) (1) = 0
[Substitute in B′ = B cos 2θ + (C - A) sin 2θ.]

C′ = (1) (1 / 2) - (-1) (12) (12) + (1) (1 / 2) = 3 / 2
[Substitute in C′ = C cos² θ - B cos θ sin θ + A sin² θ.]

D′ = (0) 12 + (0) 12 = 0
[Substitute in D′ = D cos θ + E sin θ.]

E′ = (0) 12 - (0) 12 = 0
[Substitute in E′ = E cos θ - D sin θ.]

F′ = -2
[Substitute in F′ = F.]

Hence, the equation in rotated system is (1 / 2)x2 + (0)xy + (3 / 2)y2 + (0)x + (0)y - 2 = 0
[Substitute the values.]

x2 + 3y2 = 4 or x24 + y2(43) = 1

Clearly, the above equation represents an ellipse.

9.
Choose the new coordinates of the point (- 10, 9) when the axes are rotated through an angle 30°.
 a. [, $\frac{10+9\sqrt{3}}{2}$] b. [$\frac{10\sqrt{3}-9}{2}$, $\frac{10+9\sqrt{3}}{2}$] c. [, ] d. [, $\frac{10+9\sqrt{3}}{2}$]

#### Solution:

When the axes are rotated through an angle θ then the new coordinates in terms of original coordinates are x′ = x cos θ + y sin θ and y′ = -x sin θ + y cos θ.
[Formula.]

x′ = (- 10) cos 30° + (9) sin 30°
[Substitute the values.]

= (- 10)32 + 9 / 2

= - 103+92

y′ = -(- 10) sin 30° + (9) cos 30°
[Substitute the values.]

= 10 / 2 + 932

= 10+932

New coordinates = [- 103+92, 10+932]

10.
Choose the angle to which the axes are to be rotated to remove the $\mathrm{x\text{'} y\text{'}}$ - term in the transformed equation of 9$x$2 + 3$\mathrm{xy}$ + 9$y$2 - 44 = 0.
 a. $\frac{\pi }{6}$ b. $\frac{\pi }{4}$ c. $\pi$ d. $\frac{\pi }{2}$

#### Solution:

The angle to which the axes are to be rotated to remove the x' y' - term in the transformed equation of Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is θ = 1 / 2 Cot-1[A-C / B].

A = 9, B = 3, and C = 9
[Compare with standard second degree equation.]

θ = 1 / 2Cot-1[9-93]
[Substitute the values.]

= 1 / 2Cot-1(0) = 1 / 2(π2)

= π4