88223
The transformed equation of \(3 x^{2}-6 x y+8 y^{2}=8\) when the axes are rotated about the origin through an angle \(\frac{\pi}{4}\) in the positive direction, is
1 \(5 x^{2}+10 x y+17 y^{2}+16=0\)
2 \(5 x^{2}+10 x y+17 y^{2}-16=0\)
3 \(5 x_{2}^{2}-10 x y+17 y_{2}^{2}-16=0\)
4 \(5 x^{2}-10 x y+17 y^{2}+16=0\)
Explanation:
(B) : Given, The transformation equation \(3 x^{2}-6 x y+8 y^{2}=8\) Angle of rotation about the origin \(=\frac{\pi}{4}\) We know that, \(\mathrm{X}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\therefore \quad \mathrm{X}=\mathrm{X} \cos \pi / 4-\mathrm{Y} \sin \pi / 4\) \(x=\frac{X-Y}{\sqrt{2}}\) Similarly \(y=X \sin \frac{\pi}{4}+Y \cos \frac{\pi}{4}\) \(y=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}=\frac{X+Y}{\sqrt{2}}\) Now, \(3\left(\frac{X-Y}{\sqrt{2}}\right)^{2}-6\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right)+8\left(\frac{X+Y}{\sqrt{2}}\right)^{2}=8\) Now, putting \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}\) \(\frac{3}{2}(x-y)^{2}-3\left(x^{2}-y^{2}\right)+\frac{8}{2}(x+y)^{2}=8\) \(3(x-y)^{2}-6\left(x^{2}-y^{2}\right)+8(x+y)^{2}=16\) \(3\left(x^{2}+y^{2}-2 x y\right)-6 x^{2}+6 y^{2}+8\left(x^{2}+y^{2}+2 x y\right)-16=0\) \(3 x^{2}+3 y^{2}-6 x y-6 x^{2}+6 y^{2}+8 x^{2}+8 y^{2}+16 x y-16=0\) \(5 x^{2}+17 y^{2}+10 x y-16=0\)
AP EAMCET-2018-23.04.2018
Co-Ordinate system
88224
A light ray emerging from a point source at \(A(2,3)\) is reflected on the \(y\)-axis at point ' \(B\) ' and passes through point \(\mathbf{C}(5,10)\), then the coordinates of ' \(B\) ' are
1 (5, 0)
2 (0, 5)
3 (0, 2)
4 (2, 0)
Explanation:
(B) : Given, The coordinate of source point is \(\mathrm{A}(2,3)\) The coordinate of point after reflected from Point B is C \((5,10)\) Now, Let the coordinate of point \(B\) is \((0, y)\) We know the angle of incident is equal to angle of reflection. \(\therefore\) Slope of line \(\mathrm{AB}\) from reference line \(=\) Slope of line BC from reference line \(\therefore \quad \frac{3-y}{2-0}=\frac{-10-y}{5-0}\) \(15-5 y=-20+2 y\) \(7 y=35\) \(y=5\) \(\therefore\) The coordinate of point \(\mathrm{B}\) is \((0,5)\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88225
If the point \(P\) changes to \((4,-3)\) when the axes are rotated through an angle of \(135^{\circ}\) then the coordinates of the point \(P\), with respect to the original system is
(C) : Given, The coordinate of point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) The coordinate of change \(\mathrm{P}(4,-3)\) The angle of rotated through is \(135^{\circ}\) We know that, \(\mathrm{x} =\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y} =\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\mathrm{x} =4\left(\cos 135^{\circ}\right)-(-3) \sin 135^{\circ}\) \(\mathrm{x} =4 \times\left(-\frac{1}{\sqrt{2}}\right)+3 \times \frac{1}{\sqrt{2}}\) \(\mathrm{x} =\frac{-4+3}{\sqrt{2}}\) \(\mathrm{x} =-\frac{1}{\sqrt{2}}\) Similarly, \(y=x \sin 135^{\circ}+y \cos 135\) \(=4 \times \frac{1}{\sqrt{2}}+(-3) \times\left(-\frac{1}{\sqrt{2}}\right)=\frac{7}{\sqrt{2}}\) \(\therefore\) The coordinate of point \(\mathrm{P}=\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\)
Shift-II
Co-Ordinate system
88226
The angle by which axes are to be rotated without changing the origin so that the transformed equation of \(x^{2}+4 x y-y^{2}=0\) in new coordinates \((X, Y)\) does not contain \(X Y\) term is.
88223
The transformed equation of \(3 x^{2}-6 x y+8 y^{2}=8\) when the axes are rotated about the origin through an angle \(\frac{\pi}{4}\) in the positive direction, is
1 \(5 x^{2}+10 x y+17 y^{2}+16=0\)
2 \(5 x^{2}+10 x y+17 y^{2}-16=0\)
3 \(5 x_{2}^{2}-10 x y+17 y_{2}^{2}-16=0\)
4 \(5 x^{2}-10 x y+17 y^{2}+16=0\)
Explanation:
(B) : Given, The transformation equation \(3 x^{2}-6 x y+8 y^{2}=8\) Angle of rotation about the origin \(=\frac{\pi}{4}\) We know that, \(\mathrm{X}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\therefore \quad \mathrm{X}=\mathrm{X} \cos \pi / 4-\mathrm{Y} \sin \pi / 4\) \(x=\frac{X-Y}{\sqrt{2}}\) Similarly \(y=X \sin \frac{\pi}{4}+Y \cos \frac{\pi}{4}\) \(y=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}=\frac{X+Y}{\sqrt{2}}\) Now, \(3\left(\frac{X-Y}{\sqrt{2}}\right)^{2}-6\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right)+8\left(\frac{X+Y}{\sqrt{2}}\right)^{2}=8\) Now, putting \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}\) \(\frac{3}{2}(x-y)^{2}-3\left(x^{2}-y^{2}\right)+\frac{8}{2}(x+y)^{2}=8\) \(3(x-y)^{2}-6\left(x^{2}-y^{2}\right)+8(x+y)^{2}=16\) \(3\left(x^{2}+y^{2}-2 x y\right)-6 x^{2}+6 y^{2}+8\left(x^{2}+y^{2}+2 x y\right)-16=0\) \(3 x^{2}+3 y^{2}-6 x y-6 x^{2}+6 y^{2}+8 x^{2}+8 y^{2}+16 x y-16=0\) \(5 x^{2}+17 y^{2}+10 x y-16=0\)
AP EAMCET-2018-23.04.2018
Co-Ordinate system
88224
A light ray emerging from a point source at \(A(2,3)\) is reflected on the \(y\)-axis at point ' \(B\) ' and passes through point \(\mathbf{C}(5,10)\), then the coordinates of ' \(B\) ' are
1 (5, 0)
2 (0, 5)
3 (0, 2)
4 (2, 0)
Explanation:
(B) : Given, The coordinate of source point is \(\mathrm{A}(2,3)\) The coordinate of point after reflected from Point B is C \((5,10)\) Now, Let the coordinate of point \(B\) is \((0, y)\) We know the angle of incident is equal to angle of reflection. \(\therefore\) Slope of line \(\mathrm{AB}\) from reference line \(=\) Slope of line BC from reference line \(\therefore \quad \frac{3-y}{2-0}=\frac{-10-y}{5-0}\) \(15-5 y=-20+2 y\) \(7 y=35\) \(y=5\) \(\therefore\) The coordinate of point \(\mathrm{B}\) is \((0,5)\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88225
If the point \(P\) changes to \((4,-3)\) when the axes are rotated through an angle of \(135^{\circ}\) then the coordinates of the point \(P\), with respect to the original system is
(C) : Given, The coordinate of point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) The coordinate of change \(\mathrm{P}(4,-3)\) The angle of rotated through is \(135^{\circ}\) We know that, \(\mathrm{x} =\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y} =\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\mathrm{x} =4\left(\cos 135^{\circ}\right)-(-3) \sin 135^{\circ}\) \(\mathrm{x} =4 \times\left(-\frac{1}{\sqrt{2}}\right)+3 \times \frac{1}{\sqrt{2}}\) \(\mathrm{x} =\frac{-4+3}{\sqrt{2}}\) \(\mathrm{x} =-\frac{1}{\sqrt{2}}\) Similarly, \(y=x \sin 135^{\circ}+y \cos 135\) \(=4 \times \frac{1}{\sqrt{2}}+(-3) \times\left(-\frac{1}{\sqrt{2}}\right)=\frac{7}{\sqrt{2}}\) \(\therefore\) The coordinate of point \(\mathrm{P}=\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\)
Shift-II
Co-Ordinate system
88226
The angle by which axes are to be rotated without changing the origin so that the transformed equation of \(x^{2}+4 x y-y^{2}=0\) in new coordinates \((X, Y)\) does not contain \(X Y\) term is.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88223
The transformed equation of \(3 x^{2}-6 x y+8 y^{2}=8\) when the axes are rotated about the origin through an angle \(\frac{\pi}{4}\) in the positive direction, is
1 \(5 x^{2}+10 x y+17 y^{2}+16=0\)
2 \(5 x^{2}+10 x y+17 y^{2}-16=0\)
3 \(5 x_{2}^{2}-10 x y+17 y_{2}^{2}-16=0\)
4 \(5 x^{2}-10 x y+17 y^{2}+16=0\)
Explanation:
(B) : Given, The transformation equation \(3 x^{2}-6 x y+8 y^{2}=8\) Angle of rotation about the origin \(=\frac{\pi}{4}\) We know that, \(\mathrm{X}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\therefore \quad \mathrm{X}=\mathrm{X} \cos \pi / 4-\mathrm{Y} \sin \pi / 4\) \(x=\frac{X-Y}{\sqrt{2}}\) Similarly \(y=X \sin \frac{\pi}{4}+Y \cos \frac{\pi}{4}\) \(y=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}=\frac{X+Y}{\sqrt{2}}\) Now, \(3\left(\frac{X-Y}{\sqrt{2}}\right)^{2}-6\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right)+8\left(\frac{X+Y}{\sqrt{2}}\right)^{2}=8\) Now, putting \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}\) \(\frac{3}{2}(x-y)^{2}-3\left(x^{2}-y^{2}\right)+\frac{8}{2}(x+y)^{2}=8\) \(3(x-y)^{2}-6\left(x^{2}-y^{2}\right)+8(x+y)^{2}=16\) \(3\left(x^{2}+y^{2}-2 x y\right)-6 x^{2}+6 y^{2}+8\left(x^{2}+y^{2}+2 x y\right)-16=0\) \(3 x^{2}+3 y^{2}-6 x y-6 x^{2}+6 y^{2}+8 x^{2}+8 y^{2}+16 x y-16=0\) \(5 x^{2}+17 y^{2}+10 x y-16=0\)
AP EAMCET-2018-23.04.2018
Co-Ordinate system
88224
A light ray emerging from a point source at \(A(2,3)\) is reflected on the \(y\)-axis at point ' \(B\) ' and passes through point \(\mathbf{C}(5,10)\), then the coordinates of ' \(B\) ' are
1 (5, 0)
2 (0, 5)
3 (0, 2)
4 (2, 0)
Explanation:
(B) : Given, The coordinate of source point is \(\mathrm{A}(2,3)\) The coordinate of point after reflected from Point B is C \((5,10)\) Now, Let the coordinate of point \(B\) is \((0, y)\) We know the angle of incident is equal to angle of reflection. \(\therefore\) Slope of line \(\mathrm{AB}\) from reference line \(=\) Slope of line BC from reference line \(\therefore \quad \frac{3-y}{2-0}=\frac{-10-y}{5-0}\) \(15-5 y=-20+2 y\) \(7 y=35\) \(y=5\) \(\therefore\) The coordinate of point \(\mathrm{B}\) is \((0,5)\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88225
If the point \(P\) changes to \((4,-3)\) when the axes are rotated through an angle of \(135^{\circ}\) then the coordinates of the point \(P\), with respect to the original system is
(C) : Given, The coordinate of point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) The coordinate of change \(\mathrm{P}(4,-3)\) The angle of rotated through is \(135^{\circ}\) We know that, \(\mathrm{x} =\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y} =\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\mathrm{x} =4\left(\cos 135^{\circ}\right)-(-3) \sin 135^{\circ}\) \(\mathrm{x} =4 \times\left(-\frac{1}{\sqrt{2}}\right)+3 \times \frac{1}{\sqrt{2}}\) \(\mathrm{x} =\frac{-4+3}{\sqrt{2}}\) \(\mathrm{x} =-\frac{1}{\sqrt{2}}\) Similarly, \(y=x \sin 135^{\circ}+y \cos 135\) \(=4 \times \frac{1}{\sqrt{2}}+(-3) \times\left(-\frac{1}{\sqrt{2}}\right)=\frac{7}{\sqrt{2}}\) \(\therefore\) The coordinate of point \(\mathrm{P}=\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\)
Shift-II
Co-Ordinate system
88226
The angle by which axes are to be rotated without changing the origin so that the transformed equation of \(x^{2}+4 x y-y^{2}=0\) in new coordinates \((X, Y)\) does not contain \(X Y\) term is.
88223
The transformed equation of \(3 x^{2}-6 x y+8 y^{2}=8\) when the axes are rotated about the origin through an angle \(\frac{\pi}{4}\) in the positive direction, is
1 \(5 x^{2}+10 x y+17 y^{2}+16=0\)
2 \(5 x^{2}+10 x y+17 y^{2}-16=0\)
3 \(5 x_{2}^{2}-10 x y+17 y_{2}^{2}-16=0\)
4 \(5 x^{2}-10 x y+17 y^{2}+16=0\)
Explanation:
(B) : Given, The transformation equation \(3 x^{2}-6 x y+8 y^{2}=8\) Angle of rotation about the origin \(=\frac{\pi}{4}\) We know that, \(\mathrm{X}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y}=\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\therefore \quad \mathrm{X}=\mathrm{X} \cos \pi / 4-\mathrm{Y} \sin \pi / 4\) \(x=\frac{X-Y}{\sqrt{2}}\) Similarly \(y=X \sin \frac{\pi}{4}+Y \cos \frac{\pi}{4}\) \(y=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}=\frac{X+Y}{\sqrt{2}}\) Now, \(3\left(\frac{X-Y}{\sqrt{2}}\right)^{2}-6\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right)+8\left(\frac{X+Y}{\sqrt{2}}\right)^{2}=8\) Now, putting \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}\) \(\frac{3}{2}(x-y)^{2}-3\left(x^{2}-y^{2}\right)+\frac{8}{2}(x+y)^{2}=8\) \(3(x-y)^{2}-6\left(x^{2}-y^{2}\right)+8(x+y)^{2}=16\) \(3\left(x^{2}+y^{2}-2 x y\right)-6 x^{2}+6 y^{2}+8\left(x^{2}+y^{2}+2 x y\right)-16=0\) \(3 x^{2}+3 y^{2}-6 x y-6 x^{2}+6 y^{2}+8 x^{2}+8 y^{2}+16 x y-16=0\) \(5 x^{2}+17 y^{2}+10 x y-16=0\)
AP EAMCET-2018-23.04.2018
Co-Ordinate system
88224
A light ray emerging from a point source at \(A(2,3)\) is reflected on the \(y\)-axis at point ' \(B\) ' and passes through point \(\mathbf{C}(5,10)\), then the coordinates of ' \(B\) ' are
1 (5, 0)
2 (0, 5)
3 (0, 2)
4 (2, 0)
Explanation:
(B) : Given, The coordinate of source point is \(\mathrm{A}(2,3)\) The coordinate of point after reflected from Point B is C \((5,10)\) Now, Let the coordinate of point \(B\) is \((0, y)\) We know the angle of incident is equal to angle of reflection. \(\therefore\) Slope of line \(\mathrm{AB}\) from reference line \(=\) Slope of line BC from reference line \(\therefore \quad \frac{3-y}{2-0}=\frac{-10-y}{5-0}\) \(15-5 y=-20+2 y\) \(7 y=35\) \(y=5\) \(\therefore\) The coordinate of point \(\mathrm{B}\) is \((0,5)\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88225
If the point \(P\) changes to \((4,-3)\) when the axes are rotated through an angle of \(135^{\circ}\) then the coordinates of the point \(P\), with respect to the original system is
(C) : Given, The coordinate of point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) The coordinate of change \(\mathrm{P}(4,-3)\) The angle of rotated through is \(135^{\circ}\) We know that, \(\mathrm{x} =\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta\) \(\mathrm{y} =\mathrm{X} \sin \theta+\mathrm{Y} \cos \theta\) \(\mathrm{x} =4\left(\cos 135^{\circ}\right)-(-3) \sin 135^{\circ}\) \(\mathrm{x} =4 \times\left(-\frac{1}{\sqrt{2}}\right)+3 \times \frac{1}{\sqrt{2}}\) \(\mathrm{x} =\frac{-4+3}{\sqrt{2}}\) \(\mathrm{x} =-\frac{1}{\sqrt{2}}\) Similarly, \(y=x \sin 135^{\circ}+y \cos 135\) \(=4 \times \frac{1}{\sqrt{2}}+(-3) \times\left(-\frac{1}{\sqrt{2}}\right)=\frac{7}{\sqrt{2}}\) \(\therefore\) The coordinate of point \(\mathrm{P}=\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\)
Shift-II
Co-Ordinate system
88226
The angle by which axes are to be rotated without changing the origin so that the transformed equation of \(x^{2}+4 x y-y^{2}=0\) in new coordinates \((X, Y)\) does not contain \(X Y\) term is.
