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JEE • MATHEMATICS • TRANSFORMATION OF AXES

Transformation of axes practice · Mixed difficulty

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1 Easy
1 of 38
Application +4 | -1
If first degree terms and constant term are to be removed from the equation 12x2+7xy−12y2−17x−31y−7=0,12 x^2+7 x y-12 y^2-17 x-31 y-7=0, then the origin must be shifted at the point
View answer and explanation
Correct option: A
(1,−1)(1,-1)
Option A ((1,-1)) is correct because it is the point where the origin must be shifted to remove the first-degree terms and the constant term.
2 Medium
2 of 38
Application +4 | -1
The pedal points of a perpendicular drawn from origin on the line 3x+4y−5=03 x+4 y-5=0, is
View answer and explanation
Correct option: B
(35,45)\left(\frac{3}{5}, \frac{4}{5}\right)
Option B ((3/5), (4/5)) is correct because it represents the coordinates of the foot of the perpendicular from the origin to the line.
3 Easy
3 of 38
Application +4 | -1
In order to remove xyxy-term from the equation 5x2+43  xy+9y2−8=05{x^2} + 4\sqrt 3 \;xy + 9{y^2} - 8 = 0 the coordinate axes must be rotated through an angle
View answer and explanation
Correct option: C
π/3\pi /3
Option C (π/3\pi /3) is correct because it is the angle required to rotate the coordinate axes to eliminate the xyxy term in the given equation, calculated using the formula tan⁡2θ=2ha−b\tan 2\theta = \frac{{2h}}{{a - b}}.
4 Easy
4 of 38
Application +4 | -1
For which of the following value of α=……\alpha=\ldots \ldots, the slope of the line xcos⁡α+ysin⁡α=px \cos \alpha+y \sin \alpha=p (where p≠0p \neq 0 ) is 3\sqrt{3}.
View answer and explanation
Correct option: D
5π6\frac{5 \pi}{6}
Option D (5π6\frac{5 \pi}{6}) is correct because it is the value of α\alpha for which the slope of the given line is 3\sqrt{3}, derived by using the relationship between the slope and the angle in a linear equation.
5 Medium
5 of 38
Application +4 | -1
If the origin is shifted to the point (ab/(a−b),0)(a b /(a-b), 0) without rotation, then the equation (a−b)(x2+y2)−2abx−0(a-b)\left(x^2+y^2\right)-2 a b x-0 becomes -
View answer and explanation
Correct option: D
(a−b)2(x2+y2)−a2b2(a-b)^2\left(x^2+y^2\right)-a^2 b^2
Option D ((a−b)2(x2+y2)−a2b2(a-b)^2(x^2+y^2)-a^2 b^2) is correct because shifting the origin and substituting into the equation leads to this result after simplification.
6 Easy
6 of 38
Application +4 | -1
The coordinates of the point where origin should be shifted so that the equation y2+4y+8x−2=0y^2+4 y+8 x-2=0 will not contain term in yy and the constant term, are
View answer and explanation
Correct option: A
(34,−2)\left(\frac{3}{4},-2\right)
Option A (3/4, -2) is correct because by shifting the origin to this point, the y term and the constant term vanish in the transformed equation.
7 Medium
7 of 38
Application +4 | -1
The point (4,1)(4,1) undergoes the following two successive transformation

(i) Reflection about the line y=xy=x

(ii) Translation through a distance 2 units along the positive xx-axis

Then the final coordinates of the point are
View answer and explanation
Correct option: B
(3,4)(3,4)
Option B ((3,4)) is correct because after reflection about y=x, (4,1) becomes (1,4), and then translation by 2 units along the x-axis results in (3,4).
8 Medium
8 of 38
Understanding +4 | -1
If by rotating the coordinate axes without translating the origin, the expression a1x2+2h1xy+b1y2a_1 x^2+2 h_1 x y+b_1 y^2 becomes a2X2+2h2XY+b2Y2a_2 X^2+2 h_2 X Y+b_2 Y^2, then
View answer and explanation
Correct option: B
a1+b1=a2+b2a_1+b_1=a_2+b_2
Option B is correct because a1+b1a_1 + b_1 is invariant under rotation of axes.
9 Medium
9 of 38
Application +4 | -1
If the coordinates of two points AA and BB are (3,4)(3,4) and (5,−2)(5,-2) respectively. Then, the coordinates of any point PP, if PA=PBP A=P B and area of △PAB=10\triangle P A B=10, are
View answer and explanation
Correct option: B
(7,2),(1,0)(7,2),(1,0)
Option B ((7,2),(1,0)) is correct because these points satisfy both the condition PA=PB and the area condition of the triangle PAB.
10 Easy
10 of 38
Analyzing +4 | -1
If (x,y)(x, y) and (X,Y)(X, Y) be the coordinates of the same point referred to two sets of rectangular axes with the same origin and if ax+bya x+b y becomes pX+qYp X+q Y, where a,ba, b are independent of x,yx, y, then
View answer and explanation
Correct option: B
a2+b2=p2+q2a^2+b^2=p^2+q^2
Option B is correct. By using the rotation of axes formulas and comparing coefficients, it can be shown that a2 + b2 = p2 + q2.
11 Easy
11 of 38
Application +4 | -1
If the axes are rotated through an angle of 45∘45^{\circ} in the clockwise direction, the coordinates of a point in the new system are (0,−2)(0,-2), then its original coordinates are
View answer and explanation
Correct option: D
(−2,−2)(-\sqrt{2},-\sqrt{2})
Option D (−2,−2)(-\sqrt{2},-\sqrt{2}) is correct because it is obtained by applying the rotation transformation with the given angle to the new coordinates.
12 Easy
12 of 38
Application +4 | -1
By rotating the coordinates axes through 30∘30^\circ in anticlockwise sense the equationx2+23xy−y2=2a2{x^2} + 2\sqrt 3 xy - {y^2} = 2{a^2} changes to
View answer and explanation
Correct option: B
X2−Y2=a2{X^2} - {Y^2} = {a^2}
Option B (X2−Y2=a2{X^2} - {Y^2} = {a^2}) is correct because after rotating the coordinate axes by 30 degrees and substituting the transformation equations into the original equation, the simplified equation becomes X2−Y2=a2{X^2} - {Y^2} = {a^2}.
13 Easy
13 of 38
Application +4 | -1
The transformed equation of 3x2+3y2+2xy=23{x^2} + 3{y^2} + 2xy = 2, when the coordinate axes are rotated through an angle of 45∘45^\circ, is
View answer and explanation
Correct option: B
2x2+y2=12{x^2} + {y^2} = 1
Option B (2x2+y2=12{x^2} + {y^2} = 1) is correct because after rotating the axes by 45 degrees, the transformed equation is 2x2+y2=12{x^2} + {y^2} = 1.
14 Easy
14 of 38
Application +4 | -1
The coordinates axes are rotated through an angle 135∘135^\circ. If the coordinates of a point PP in the new system are known to be (4,  −3)\left( {4,\; - 3} \right), then the coordinates of PP in the original system are
View answer and explanation
Correct option: D
(−12,72)\left( { - \frac{1}{{\sqrt 2 }},\frac{7}{{\sqrt 2 }}} \right)
Option D (−12,72)\left( { - \frac{1}{{\sqrt 2 }},\frac{7}{{\sqrt 2 }}} \right) is correct because it represents the original coordinates of point P after rotating the axes by 135 degrees.
15 Easy
15 of 38
Application +4 | -1
If pp and p′p^{\prime} are the lengths of perpendiculars drawn from the origin to lines xsec⁡θ+ycosec⁡θ=2ax \sec \theta+y \operatorname{cosec} \theta=2 a and xcos⁡θ+ysin⁡θ=acos⁡2θx \cos \theta+y \sin \theta=a \cos 2 \theta respectively, then (pp′+p′p)2=……\left(\frac{p}{p^{\prime}}+\frac{p^{\prime}}{p}\right)^2=\ldots \ldots
View answer and explanation
Correct option: B
4cosec⁡24θ4 \operatorname{cosec}^2 4 \theta
Option B (4cosec⁡24θ4 \operatorname{cosec}^2 4 \theta) is correct because it represents the simplified form of the expression (pp′+p′p)2\left(\frac{p}{p^{\prime}}+\frac{p^{\prime}}{p}\right)^2 after substituting the values of pp and p′p^{\prime} and using trigonometric identities.

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16 Easy
16 of 38
Application +4 | -1
If we shift the origin to (−2,3)(-2,3) then will become (3,−2)(3,-2).
View answer and explanation
Correct option: B
(1,1)(1,1)
Option B ((1,1)(1,1)) is correct because it represents the new coordinates after shifting the origin as specified.
17 Easy
17 of 38
Application +4 | -1
The cartesian form of the equation r1/2=a1/2cos⁡θ2r^{1 / 2}=a^{1 / 2} \cos \frac{\theta}{2} is
View answer and explanation
Correct option: A
(2x2+2y2−ax)2=a2(x2+y2)\left(2 x^2+2 y^2-a x\right)^2=a^2\left(x^2+y^2\right)
Option A ((2x2+2y2−ax)2=a2(x2+y2)2x^2+2y^2-ax)^2=a^2(x^2+y^2)) is correct because it is the cartesian form derived from the given polar equation through trigonometric identities and coordinate transformations.
18 Easy
18 of 38
Application +4 | -1
If by shifting the origin at (1,  1)\left( {1,\;1} \right) the coordinates of a point PP become (cos⁡θ,cos⁡ϕ),\left( {\cos \theta ,\cos \phi } \right), then the original coordinates of PP were
View answer and explanation
Correct option: A
(2cos⁡2θ/2,  2cos⁡2ϕ/2)\left( {2{{\cos }^2}\mathrm{\theta }/2,\;2{{\cos }^2}\phi /2} \right)
Option A (2cos⁡2θ/2,  2cos⁡2ϕ/2)\left( {2{{\cos }^2}\mathrm{\theta }/2,\;2{{\cos }^2}\phi /2} \right) is correct because it represents the original coordinates of P after shifting the origin back, using the identity 1+cos⁡θ=2cos⁡2(θ/2)1 + \cos \theta = 2{\cos ^2}(\theta /2).
19 Easy
19 of 38
Application +4 | -1
Keeping the origin constant axes are rotated at an angle 30∘30^{\circ} in negative direction, then coordinates of (2,1)(2,1) with respect to new axes are
View answer and explanation
Correct option: D
None of these
The correct answer is None of these. The calculation of X' and Y' after rotation is incorrect in the provided solution.
20 Easy
20 of 38
Application +4 | -1
If the axes are shifted to the point (1,−2)(1,-2) without rotation, what does the equation 2x2+y2−4x+4y=02 x^2+y^2-4 x+4 y=0 become?
View answer and explanation
Correct option: B
2X2+Y2=62 X^2+Y^2=6
Option B (2X² + Y² = 6) is correct because substituting x = X+1 and y = Y-2 into the original equation and simplifying results in 2X² + Y² = 6.
21 Medium
21 of 38
Analyzing +4 | -1
If by rotating the coordinate axes without translating the origin, the expression a1x2+2h1xy+b1y2a_1 x^2+2 h_1 x y+b_1 y^2 becomes a2X2+2h2XY+b2Y2a_2 X^2+2 h_2 X Y+b_2 Y^2, then
View answer and explanation
Correct option: A
(a1−b1)2+4h12=(a2−b2)2+4h22\left(a_1-b_1\right)^2+4 h_1^2=\left(a_2-b_2\right)^2+4 h_2^2
Option A ((a1−b1)2+4h12=(a2−b2)2+4h22)\left((a_1 - b_1)^2 + 4h_1^2 = (a_2 - b_2)^2 + 4h_2^2\right) is correct because it represents an invariant under rotation of coordinate axes.
22 Easy
22 of 38
Application +4 | -1
The angle through which the coordinates axes be rotated so that xyx y-term in the equation 5x2+43xy+9y2=05 x^2+4 \sqrt{3} x y+9 y^2=0 may be missing, is
View answer and explanation
Correct option: C
π/3\pi / 3
Option C (π/3\pi / 3) is correct because it is the angle that eliminates the xyxy term when the axes are rotated, as found by solving the equation tan⁡2θ=2ha−b\tan 2\theta = \frac{2h}{a-b}.
23 Easy
23 of 38
Application +4 | -1
If the axes be turned through an angle tan⁡−12\tan ^{-1} 2, then the equation 4xy−3x2=a24 x y-3 x^2=a^2 transforms to
View answer and explanation
Correct option: C
X2−4Y2=a2X^2-4 Y^2=a^2
Option C (X² - 4Y² = a²) is correct because rotating the axes through the given angle transforms the equation to X² - 4Y² = a².
24 Easy
24 of 38
Application +4 | -1
If the axes are transformed from origin to the point (3,4)(3,4), then new coordinates of (7,8)(7,8) are
View answer and explanation
Correct option: C
(4,4)(4,4)
Option C (4,4) is correct because it is found by subtracting the old origin coordinates from the point's coordinates.
25 Medium
25 of 38
Application +4 | -1
ABCA B C is a variable triangle with the fixed vertex C(1,2)C(1,2) and A,BA, B having the coordinates (cos⁡t,sin⁡t),(sin⁡t,−cos⁡t)(\cos t, \sin t),(\sin t,-\cos t) respectively, where tt is a parameter. The locus of centroid of the △ABC\triangle A B C is
View answer and explanation
Correct option: B
3(x2+y2)−2x−4y+1=03\left(x^2+y^2\right)-2 x-4 y+1=0
Option B (3(x2+y2)−2x−4y+1=03(x^2+y^2)-2x-4y+1=0) is correct because it represents the locus of the centroid of the triangle ABC, derived by eliminating the parameter t.
26 Medium
26 of 38
Application +4 | -1
The transformed equation of x2+6xy+8y2=10{x^2} + 6xy + 8{y^2} = 10 when the axes are rotated through an angle π4\frac{\pi }{4} is
View answer and explanation
Correct option: C
15x2+14xy+3y2=2015{x^2} + 14xy + 3{y^2} = 20
Option C (15x2+14xy+3y2=2015{x^2} + 14xy + 3{y^2} = 20) is correct because after rotating the axes by π4\frac{\pi}{4}, the transformed equation is 15x2+14xy+3y2=2015{x^2} + 14xy + 3{y^2} = 20.
27 Easy
27 of 38
Application +4 | -1
Without change of axes the origin is shifted to (h,  k)\left( {h,\;k} \right), then from the equation

x2+y2−4x+6y−7=0{x^2} + {y^2} - 4x + 6y - 7 = 0 the terms containing linear powers are missing. The point (h,  k)\left( {h,\;k} \right) is
View answer and explanation
Correct option: C
(2,−3)\left( {2, - 3} \right)
Option C ((2, -3)) is correct because shifting the origin to this point eliminates the linear terms in the equation.
28 Easy
28 of 38
Application +4 | -1
If origin is shifted to (7,−4)\left( {7, - 4} \right), then point (4, 5) shifted to
View answer and explanation
Correct option: A
(−3,9)\left( { - 3,9} \right)
Option A ((-3, 9)) is correct because shifting the origin involves subtracting the new origin's coordinates from the original point's coordinates.
29 Easy
29 of 38
Application +4 | -1
To remove the first degree terms in the equation 4x2+9y2−8x+36y+4=04 x^2+9 y^2-8 x+36 y+4=0, the origin in shifted to the point
View answer and explanation
Correct option: B
(1,−2)(1,-2)
Option B (1,−2)(1,-2) is correct because it represents the point to which the origin must be shifted to eliminate the first-degree terms in the given equation.
30 Easy
30 of 38
Understanding +4 | -1
If by rotating the coordinate axes without translating the origin, the expression a1x2+2h1xy+b1y2a_1 x^2+2 h_1 x y+b_1 y^2 becomes a2X2+2h2XY+b2Y2a_2 X^2+2 h_2 X Y+b_2 Y^2, then
View answer and explanation
Correct option: C
a1b1−h12=a2b2−h22a_1 b_1-h_1^2=a_2 b_2-h_2^2
Option C is correct because a1b1−h12a_1b_1 - h_1^2 is invariant under rotation of axes.
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