Class 11 Mathematics Transformation of Axes MCQs Set 01

Question. If (2, 3) are the coordinates of a point P in the new system when the origin is shifted to (-3, 7) then the original coordinates of P are
(a) (-1, 10)
(b) (5, -4)
(c) (-5, 4)
(d) (-1, 5)
Answer: (a) (-1, 10)

Question. The coordinates of the point (4,5) in the new system, when its origin is shifted to (3,7) are
(a) (1, 2)
(b) (-1, 2)
(c) (-1, -2)
(d) (1, -2)
Answer: (d) (1, -2)

Question. When the origin is shifted to a point P, the point (2, 0) is transformed to (0, 4) then the coordinates of P are
(a) (2, -4)
(b) (-2, 4)
(c) (-2, -4)
(d) (2, 4)
Answer: (a) (2, -4)

Question. If the axes are translated to the point (-2,-3) then the equation \( x^2 + 3y^2 + 4x + 18y + 30 = 0 \) transforms to
(a) \( X^2 + Y^2 = 4 \)
(b) \( X^2 + 3Y^2 = 1 \)
(c) \( X^2 - Y^2 = 4 \)
(d) \( X^2 - 3Y^2 = 1 \)
Answer: (b) \( X^2 + 3Y^2 = 1 \)

Question. When the axes are translated to the point (5, -2) then the transformed form of the equation \( xy + 2x - 5y - 11 = 0 \) is
(a) \( \frac{X}{Y} = 1 \)
(b) \( \frac{Y}{X} = 1 \)
(c) \( XY = 1 \)
(d) \( XY^2 = 2 \)
Answer: (c) \( XY = 1 \)

Question. If the transformed equation of a curve when the origin is translated to (1,1) is \( X^2 + Y^2 + 2X - Y + 2 = 0 \) then the original equation of the curve is
(a) \( x^2 + 2y^2 = 1 \)
(b) \( x^2 + y^2 + 3y + 3 = 0 \)
(c) \( x^2 + y^2 + 3y - 3 = 0 \)
(d) \( x^2 + y^2 - 3y + 3 = 0 \)
Answer: (d) \( x^2 + y^2 - 3y + 3 = 0 \)

Question. In order to make the first degree terms missing in the equation \( 2x^2 + 7y^2 + 8x - 14y + 15 = 0 \), the origin should be shifted to the point
(a) (1, -2)
(b) (-2, -1)
(c) (2, 1)
(d) (-2, 1)
Answer: (d) (-2, 1)

Question. The point to which the origin should be shifted in order to remove the x and y terms in the equation \( 14x^2 - 4xy + 11y^2 - 36x + 48y + 41 = 0 \) is
(a) (1, -2)
(b) (-2, 1)
(c) (-1, 2)
(d) (2, -1)
Answer: (a) (1, -2)

Question. If the distance between the two given points is 2 units and the points are transferred by shifting the origin to (2, 2), then the distance between the points in their new position is
(a) 2
(b) 5
(c) 6
(d) 7
Answer: (a) 2

Question. When (0, 0) shifted to (3, -3) the coordinates of P(5, 5), Q(-2, 4) and R(7, -7) in the new system are A, B, C then area of triangle ABC in sq units is
(a) 43
(b) 23
(c) 45
(d) 50
Answer: (a) 43

Question. When axes are rotated through an angle of \( 45^o \) in positive direction without changing origin then the coordinates of \( (\sqrt{2}, 4) \) in old system are
(a) \( (1-2\sqrt{2}, 1+2\sqrt{2}) \)
(b) \( (1+2\sqrt{2}, 1-2\sqrt{2}) \)
(c) \( (2\sqrt{2}, \sqrt{2}) \)
(d) \( (2, \sqrt{2}) \)
Answer: (a) \( (1-2\sqrt{2}, 1+2\sqrt{2}) \)

Question. If the axes are rotated through an angle \( 30^o \), the coordinates of \( (2\sqrt{3}, -3) \) in the new system are
(a) \( \left( \frac{3}{2}, \frac{-5}{2} \right) \)
(b) \( \left( \frac{-\sqrt{3}}{2}, \frac{5}{2} \right) \)
(c) \( \left( \frac{3-5\sqrt{3}}{2}, \frac{3}{2} \right) \)
(d) \( \left( 3\sqrt{2}, \frac{-5\sqrt{3}}{2} \right) \)
Answer: (c) \( \left( \frac{3-5\sqrt{3}}{2}, \frac{3}{2} \right) \)

Question. The transformed equation of \( x^2 + 2\sqrt{3}xy - y^2 - 8 = 0 \), when the axes are rotated through an angle \( \frac{\pi}{6} \) is
(a) \( X^2 - Y^2 = 0 \)
(b) \( X^2 - Y^2 = 4 \)
(c) \( X^2 - Y^2 = 2 \)
(d) \( X^2 + Y^2 = 4 \)
Answer: (b) \( X^2 - Y^2 = 4 \)

Question. If the axes are rotated through an angle \( 180^o \) then the equation \( 2x - 3y + 4 = 0 \) becomes
(a) \( 2X - 3Y - 4 = 0 \)
(b) \( 2X + 3Y - 4 = 0 \)
(c) \( 3X - 2Y + 4 = 0 \)
(d) \( 3X + 2Y + 4 = 0 \)
Answer: (a) \( 2X - 3Y - 4 = 0 \)

Question. If the transformed equation of a curve is \( 17X^2 - 16XY + 17Y^2 = 225 \) when the axes are rotated through an angle \( 45^o \), then the original equation of the curve is
(a) \( 25x^2 + 9y^2 = 225 \)
(b) \( 9x^2 + 25y^2 = 225 \)
(c) \( 25x^2 - 9y^2 = 225 \)
(d) \( 9x^2 - 25y^2 = 225 \)
Answer: (a) \( 25x^2 + 9y^2 = 225 \)

Question. If the equation \( 4x^2 + 2\sqrt{3}xy + 2y^2 - 1 = 0 \) becomes \( 5X^2 + Y^2 = 1 \), when the axes are rotated through an angle \( \theta \), then \( \theta \) is
(a) \( 15^o \)
(b) \( 30^o \)
(c) \( 45^o \)
(d) \( 60^o \)
Answer: (b) \( 30^o \)

Question. The angle of rotation of axes in order to eliminate \( xy \) term in the equation \( xy = c^2 \) is
(a) \( \frac{\pi}{12} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{4} \)
Answer: (d) \( \frac{\pi}{4} \)

Question. The transformed equation of \( x^2 + y^2 = r^2 \), when the axes are rotated through an angle \( 36^o \) is
(a) \( \sqrt{5}X^2 - 4XY + Y^2 = r^2 \)
(b) \( X^2 + 2XY - \sqrt{5}Y^2 = r^2 \)
(c) \( X^2 - Y^2 = r^2 \)
(d) \( X^2 + Y^2 = r^2 \)
Answer: (d) \( X^2 + Y^2 = r^2 \)

Question. To remove the first degree terms of the equation \( 2xy + 4x - 2y + 7 = 0 \) the shifted origin is
(a) (2, -1)
(b) (-1, 2)
(c) (1, -2)
(d) (-2, 1)
Answer: (c) (1, -2)

Question. By translating the axes the equation \( xy - 2x - 3y - 4 = 0 \) has changed to \( XY = k \), then \( k = \)
(a) -10
(b) 10
(c) 4
(d) -4
Answer: (b) 10

Question. The origin is shifted to (1, 2), the equation \( y^2 - 8x - 4y + 12 = 0 \) changes to \( Y^2 + 4aX = 0 \) then \( a = \)
(a) 2
(b) -2
(c) 1
(d) -1
Answer: (b) -2

Question. When the origin is shifted to a suitable point, the equation \( 2x^2 + y^2 - 4x + 4y = 0 \) transformed as \( 2X^2 + Y^2 - 8X + 8Y + 18 = 0 \). The point to which origin was shifted is
(a) (1, 2)
(b) (1, -2)
(c) (-1, 2)
(d) (-1, -2)
Answer: (c) (-1, 2)

Question. If \( (\cos\alpha, \cos\beta) \) are the new co-ordinates of a point P when the axes are translated to the point (1,1), then the original coordinates are
(a) \( (2\cos^2\alpha/2, 2\cos^2\beta/2) \)
(b) \( (2\cos^2\alpha/2, 2\sin^2\beta/2) \)
(c) \( (2\sin^2\alpha/2, 2\cos^2\beta/2) \)
(d) \( (-2\cos^2\alpha/2, -2\cos^2\beta/2) \)
Answer: (a) \( (2\cos^2\alpha/2, 2\cos^2\beta/2) \)

Question. The first degree terms of \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) are removed by shifting origin to \( (\alpha, \beta) \). The new equation is
(a) \( ax^2 + 2hxy + by^2 + 2y\alpha + 2b\beta + c = 0 \)
(b) \( ax^2 + 2hxy + by^2 + g\alpha + f\beta + c = 0 \)
(c) \( ax^2 + 2hxy + by^2 + h\alpha + b\beta + c = 0 \)
(d) \( ax^2 + 2hxy + by^2 - h\alpha - b\beta - c = 0 \)
Answer: (b) \( ax^2 + 2hxy + by^2 + g\alpha + f\beta + c = 0 \)

Question. When (0, 0) shifted to (2, -2) the transformed equation of \( (x-2)^2 + (y+2)^2 = 9 \) is
(a) \( X^2 + Y^2 = 9 \)
(b) \( X^2 + 3Y^2 = 9 \)
(c) \( X^2 + Y^2 - 2X + 6Y = 0 \)
(d) \( 4X^2 + 9Y^2 = 36 \)
Answer: (a) \( X^2 + Y^2 = 9 \)
 

Question. The transformed equation of \( 4xy - 3x^2 = 10 \) when the axes are rotated through an angle whose tangent is '2' is
(a) \( X^2 - 4Y^2 = 10 \)
(b) \( 4X^2 - Y^2 = 10 \)
(c) \( XY - 10 = 0 \)
(d) \( 2X^2 - Y^2 + 10 = 0 \)
Answer: (a) \( X^2 - 4Y^2 = 10 \)

Question. The angle of rotation of the axes so that the equation \( \sqrt{3}x - y + 5 = 0 \) may be reduced to the form \( Y = k \), where \( k \) is a constant is
(a) \( \pi/6 \)
(b) \( \pi/4 \)
(c) \( \pi/3 \)
(d) \( \pi/12 \)
Answer: (c) \( \pi/3 \)

Question. The angle of rotation of the axes so that the equation \( ax + by + c = 0 \) may be reduced to \( X = p \) is
(a) \( \tan^{-1} \frac{b}{a} \)
(b) \( \tan^{-1} \frac{a}{b} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{3} \)
Answer: (a) \( \tan^{-1} \frac{b}{a} \)

Question. The coordinate axes are rotated about the origin 'O' in the counter clockwise direction through an angle \( 60^o \). If \( a \) and \( b \) are the intercepts made on the new axes by a straight line whose equation referred to the original axes is \( 3x + 4y - 5 = 0 \) then \( \frac{1}{a^2} + \frac{1}{b^2} = \)
(a) 1/25
(b) 1/9
(c) 1/16
(d) 1
Answer: (d) 1

Question. The coordinate axes are rotated through an angle \( \theta \) about the origin in anticlock-wise sense. If the equation \( 2x^2 + 3xy - 6x + 2y - 4 = 0 \) changes to \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) then \( a+b \) is equal to
(a) \( 3\cos\theta - 3\sin\theta \)
(b) \( 3\cos\theta + 2\sin\theta \)
(c) 1
(d) 2
Answer: (d) 2

Question. Let L be the line \( 2x+y-2=0 \). The axes are rotated by \( 45^o \) in clockwise direction then the intercepts made by the line L on the new axes are respectively
(a) \( 1, \sqrt{2} \)
(b) \( \sqrt{2}, 1 \)
(c) \( 2\sqrt{2}, \frac{2\sqrt{2}}{3} \)
(d) \( \frac{2\sqrt{2}}{3}, 2\sqrt{2} \)
Answer: (c) \( 2\sqrt{2}, \frac{2\sqrt{2}}{3} \)

Question. A point (2,2) undergoes reflection in the x-axis and then the coordinate axes are rotated through an angle of \( \pi/4 \) in anticlockwise direction. The final position of the point in the new coordinate system is
(a) \( (0, 2\sqrt{2}) \)
(b) \( (0, -2\sqrt{2}) \)
(c) \( (2\sqrt{2}, 0) \)
(d) \( (-2\sqrt{2}, 0) \)
Answer: (b) \( (0, -2\sqrt{2}) \)

Question. The acute angle \( \theta \) through which the coordinate axes should be rotated for the point A (2,4) to attain the new abscissa 4 is given by
(a) \( \tan\theta = 3/4 \)
(b) \( \tan\theta = 5/6 \)
(c) \( \tan\theta = 7/8 \)
(d) \( \tan\theta = \frac{3}{2} \)
Answer: (a) \( \tan\theta = 3/4 \)

Question. A line has intercepts \( a, b \) on the axes when the axes are rotated through an angle \( \alpha \), the line makes equal intercepts on axes then \( \tan\alpha = \)
(a) \( \frac{a+b}{a-b} \)
(b) \( \frac{a-b}{a+b} \)
(c) \( \frac{a}{b} \)
(d) \( \frac{b}{a} \)
Answer: (b) \( \frac{a-b}{a+b} \)

Question. The new equation of the curve \( 4(x-2y+1)^2 + 9(2x+y+2)^2 = 25 \), if the lines \( 2x+y+2=0 \) and \( x-2y+1=0 \) are taken as the new x and y axes respectively is
(a) \( 4X^2 + 9Y^2 = 5 \)
(b) \( 4X^2 + 9Y^2 = 25 \)
(c) \( 4X^2 + 9Y^2 = 7 \)
(d) \( 4X^2 - 9Y^2 = 7 \)
Answer: (a) \( 4X^2 + 9Y^2 = 5 \)

Question. The line joining the points A(2,0) and B(3,1) is rotated through an angle of \( 45^o \), about A in the anticlock wise direction. the coordinates of B in the new position (EAM-2011)
(a) \( (2, \sqrt{2}) \)
(b) \( (\sqrt{2}, 2) \)
(c) (2,2)
(d) \( (\sqrt{2}, \sqrt{2}) \)
Answer: (a) \( (2, \sqrt{2}) \)

Assertion and Reason Questions

Question. Assertion(A): If the area of triangle formed by (0,0), (2, 0), (0, 2) is 2 square units. Then the area of triangle on shifting the origin to a point (2,3) is 2 units
Reason(R): By the change of axes area does not change
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true and R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is false
Answer: (a) Both A and R are true and R is the correct explanation of A

Question. Statement I : The point to which the origin has to be shifted to eliminate x and y terms in the equation \( a(x+\alpha)^2 + b(y+\beta)^2 = c \) is \( (-\alpha, -\beta) \)
Statement II : The point to which the origin has to be shifted to eliminate x and y terms in \( ax^2 + by^2 + 2gx + 2fy + c = 0 \) is \( \left( -\frac{g}{a}, -\frac{f}{b} \right) \)
Which of the above statement is true :
(a) only I
(b) only II
(c) Both I and II
(d) Neither I nor II
Answer: (c) Both I and II

Question. Statement-1: By translating the axes the equation \( xy - x + 2y = 6 \) has changed to \( XY = c \) then c=4
Statement-2: If the axes are translated to the point \( (h, k) \), then the equation \( f(x, y) = 0 \) of a curve is transformed to \( f(X-h, Y-k) = 0 \)
(a) Statement-1: is true, statement-2 true, statement-2 is a correct explanation for statement-1
(b) Statement-1: is true, statement-2 true, statement-2 is not a correct explanation for statement-1
(c) Statement-1 is true, statement-2 is false
(d) statement-1 is false, statement-2 is true
Answer: (c) Statement-1 is true, statement-2 is false

Question. To remove the first degree terms in the following equations origin should be shifted to the another point then calculate the new origins for List - II
List - I
(A) \( x^2 - y^2 + 2x + 4y = 0 \)
(B) \( 4x^2 + 9y^2 - 8x + 36y + 4 = 0 \)
(C) \( x^2 + 3y^2 - 2x + 12y + 1 = 0 \)
(D) \( 2(x - 5)^2 + 3(y + 7)^2 = 10 \)
List - II
1) (5, -7)
2) (1, -2)
3) (-1, 2)
4) (-1, -2)
5) (-5, 7)
The correct matching is:
(a) A-4, B-2, C-2, D-5
(b) A-5, B-3, C-3, D-5
(c) A-3, B-2, C-2, D-1
(d) A-4, B-3, C-3, D-1
Answer: (c) A-3, B-2, C-2, D-1