# Shifting of Axes

## Trending Questions

**Q.**The transformed equation of x2+2y2+2x−4y+2=0 when origin is shifted to the point (−1, 1) with out rotation of axes, is

- X2−2Y2=1
- 2X2+Y2=1
- X2+2Y2=1
- X2−Y2=0

**Q.**If the origin is shifted to (7, -4), then the coordinates of the point P(4, 5) will become

- (-3, 9)
- (-3, -9)
- (3, 9)
- (3, -9)

**Q.**

Write the distance between the vertex and focus of the parabola y2+6y+2x+5=0.

**Q.**

Find what the following equations become when the origin is shifted to the point(1, 1)?(i) x2+xy−3y2−y+2=0(ii) xy−y2−x+y=0(iii) xy−x−y+1=0(iv) x2−y2−2x+2y=0

**Q.**

Axis of a parabola is $y=x$ and vertex and focus are at a distance $\sqrt{2}$ and $2\sqrt{2}$, respectively from the origin. Then, equation of the parabola is

${(x\xe2\u20ac\u201cy)}^{2}=8(x+y\xe2\u20ac\u201c2)$

${(x+y)}^{2}=2(x+y\xe2\u20ac\u201c2)$

${(x\xe2\u20ac\u201cy)}^{2}=4(x+y\xe2\u20ac\u201c2)$

${(x+y)}^{2}=2(x-y+2)$

**Q.**

A rod of length I sides with its ends on two perpendicular lines. Then, the locus of its midpoint is

**Q.**A cube of side 3 units has one vertex at point (1, 1, 1) and the three edges from this vertex are respectively parallel to positive x - axis and negative y and z - axes. Find the coordinates of other vertices of the cube.

**Q.**The transformed coordinates of the point (4, 3) when the axes are translated to point (3, 1) and then rotated through 30∘ in anticlockwise direction is

- (2√3+12, √3−22)
- (√3+12, 2√3+12)
- (√3+22, 2√3−12)
- (√3−22, √3+12)

**Q.**

Find the area of the region bounded by
*y*^{2} = 9*x*, * x* = 2, *x* = 4 and the
*x*-axis in the first quadrant.

**Q.**

Find the area of the region bounded by
*x*^{2} = 4*y*, *y* = 2, *y* = 4 and the
*y*-axis in the first quadrant.

**Q.**

verify that the area of the triangle with vertices (2, 3) , (5, 7) and (-3-1) aremains invariant under the translation of axes when the origin is shifted to the point(-1, 3).

**Q.**By shifting origin to a suitable point with out rotation of axes, the equation xy−x+2y=6 has transformed to XY=B, then the value of B is

**Q.**x - axis is a tangent and y - axis is normal to a parabola whose focus is (2, 3)

The equation of tangent at vertex of parabola is

- 2x - 3y+9 = 0
- 2x - 3y - 4 = 0
- 3x + 2y - 6 = 0
- 3x + 2y + 1 = 0

**Q.**

Verify that area of the triangle with vertices (4, 6) (7, 10) and(1, -2) remains invariant under the translation of axes when the origin is shifted to the point (-2, 1).

**Q.**If (7, 5) are the new coordinates of P when origin is shifted to (1, 1), then the original coordinates of P are

- (8, 6)
- (6, 4)
- (−6, −4)
- (4, 3)

**Q.**

In the new coordinate system origin is shifted to (h, k) and the axes are rotated through angle of 90∘ in the anti-clockwise direction. The new co-ordinates of (x, y) is obtained by the following method.

1) (x + iy) becomes (x + iy) e−iπ2

Let it be (x+iy) or (x, y)

2)(x, y) becomes (x - h, y - k)

True

False

**Q.**

What does the equation (a−b)(x2+y2)−2 abx=0 become if the origin is shifted to the point (aba−b, 0) without rotation ?

**Q.**

Find the point to which the origin should be dhifted after a translation of axes so that the following equations will have no first degree terms:?(i) y2+x2−4x−8y+3=0(ii) x2+y2−5x+2y−5=0(iii) x2−12x+4=0

**Q.**

The number of straight lines equally inclined to both the axes are

A) 1

B) 0

C) 2

D) Infinite

**Q.**

At what point the origin be shifted so that the equationx2+xy−3x−y+2=0does not contain any first degree term and constant term ?

**Q.**

Write the equation of the direction of the parabola x2−4x−8y+12=0.

**Q.**The point where the origin has to be shifted so that the equation y2+4y+8x−2=0 will not contain y and the constant term is

- (34, −2)
- (35, −2)
- (2, 4)
- (1, −4)

**Q.**

Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree)

terms in the equation x2+y2−4x+6y−7=0 are eliminated. Then the point (h, k) is

(3, 2)

(- 3, 2)

(2, - 3)

(1.7)

**Q.**Equation of the image of the pair of rays y = | x | by the line x = 1 is

- | y | + 2 = x
- none of these
- y = | x - 2|
- | y | = x + 2

**Q.**When the origin is shifted to (1, 2), the equation y2−8x−4y+12=0 changes to Y2=4aX, then value of a is

**Q.**The point where the origin has to be shifted so that the equation y2+4y+8x−2=0 will not contain y and the constant term is

- (34, −2)
- (35, −2)
- (2, 4)
- (1, −4)

**Q.**

The new coordinates of a point (4, 5), when the origin is shifted to the point (1, -2) are

(3, 7)

None of these

(3, 5)

(5, 3)

**Q.**

If origin is shifted to the point (2, 3) without rotation of axes then the coordinates of the point P which divides the join of A(4, 8) and B(7, 14) in the ratio 1:2 or 2:1 with respect to the new system of coordinates can be

(2, 5)

(3, 7)

(4, 9)

(5, 11)

**Q.**

Find what the following equations become when the origin is shifted to the point(1, 1)?(i) x2+xy−3y−y+2=0(ii) x2−y2−2x+2y=0(iii) xy−x−y+1=0(iv) xy−y2−x+y=0

**Q.**If the origin is shifted to (7, -4), then the coordinates of the point P(4, 5) will become

- (-3, 9)
- (-3, -9)
- (3, 9)
- (3, -9)