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Co-Ordinate system

88214 Find the transformed equation of $x \cos θ + y$ $\sin θ = p$ , when the axes are rotated through an angle $θ$ .

1

x = p

2

y = p

3

x + y = p

4

x - y = p

Explanation:

(A) : Given,
The transformed equation is,
$x \cos θ + y \sin θ = p$
The axis of rotation $= θ$
We know that,
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Now, in equation (i) -
$x \cos θ + y \sin θ = p$
$(x^{'} \cos θ - y^{'} \sin θ) \cos θ + (x^{'} \sin θ + y^{'} \cos θ) \sin θ = p$
$x^{'} \cos^{2} θ - y^{'} \sin θ \cos θ + x^{'} \sin^{2} θ + y^{'} \sin θ \cos θ = p$
$x^{'} [\cos^{2} θ + \sin^{2} θ] = p$
$x^{'} = p$
$x = p$

AP EAMCET-2020-17.09.2020

Co-Ordinate system

88215 When the origin is shifted to $(2, 3)$ the transformed equation $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0$ , then the original equation of curve is .......

1

x^{2} - 2 y^{2} - 3 x y + 4 x - y + 20 = 0

2

x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0

3

x^{2} - 2 y^{2} - 3 x y - 4 x - y + 20 = 0

4

x^{2} - 2 y^{2} - 3 x y + 4 x - y - 20 = 0

Explanation:

(B) : Given,
The coordinate of shifted point is $(2, 3)$
The equation is $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0 \dots$ (i)
To get the original equation of curve, replace $(x, y)$ by
$(x - 2, y - 3)$
Now, Putting the value of $x$ and $y$ in equation (i)
$(x - 2)^{2} + 3 (x - 2) (y - 3) - 2 (y - 3)^{2} + 17 (x - 2) - 7 (y - 3)$ $- 11 = 0$
$x^{2} + 4 - 4 x + 3 (x y - 3 x - 2 y + 6) - 2 (y^{2} + 9 - 6 y) + 17 x$ $- 34 - 7 y + 21 - 11 = 0$
$x^{2} + 4 - 4 x + 3 x y - 9 x - 6 y + 18 - 2 y^{2} - 18 + 12 y +$ $17 x - 34 - 7 y + 21 - 11 = 0$
$x^{2} + 3 xy - 2 y^{2} + 4 x - y - 20 = 0$
$x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0$

AP EAMCET-2020-18.09.2020

Co-Ordinate system

88216 The polar equation of the line perpendicular to the line $\sin θ - \cos θ = \frac{1}{r}$ and passing through the point $(2, \frac{π}{6})$ is

1

\sin θ + \cos θ = \frac{\sqrt{3} + 1}{r}

2

\sin θ - \cos θ = \frac{\sqrt{3} + 1}{r}

3

\sin θ + \cos θ = \frac{\sqrt{3} - 1}{r}

4

\cos θ - \sin θ = \frac{\sqrt{3}}{r}

Explanation:

(A) : Given,
The polar equation is, $\sin θ - \cos θ = \frac{1}{r}$
$∴ r \sin θ - r \cos θ = 1$
The perpendicular equation passing through
the point $(2, π / 6)$
We know that,
The conversion of polar coordinate into cartesian coordinate
$x = r \cos θ$
$y = r \sin θ$
$∴$ From equation (i)
$y - x = 1$
The slope of equation (ii) is
$m_{1} = 1$
The slope of perpendicular equation $m_{2} = - 1$
Now, The equation of line which passes through point $(2, π / 6)$
$(y - y_{1}) = m_{2} (x - x_{1})$
$(r \sin θ - 2 \sin π / 6) = - 1 (r \cos θ - 2 \cos π / 6)$
$(r \sin θ - 2 \times \frac{1}{2}) = - 1 (r \cos θ - 2 \times \frac{\sqrt{3}}{2})$
$r \sin θ - 1 = - r \cos θ + \sqrt{3}$
$r (\sin θ + \cos θ) = 1 + \sqrt{3}$
$\sin θ + \cos θ = \frac{1 + \sqrt{3}}{r}$

AP EAMCET-2011

Co-Ordinate system

88217 The origin is translated to $(1, 2)$ . The point $(7$ , 5) in the old system undergoes the following transformations successively.
I. Moves to the new point under the given translation of origin.
II. Translated through 2 units along the negative direction of the new $X$ -axis.
III. Rotated through an angle $\frac{π}{4}$ about the origin of new system in the clockwise direction. The final position of the point $(7, 5)$ is

1

(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

2

(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

(C) : Given,
The old coordinate of point $P = (7, 5)$
The coordinate shifted by $(1, 2)$
We know that,
The new coordinate after translation
$P^{'} = (7 - 1, 5 - 2)$
$P^{'} = (6, 3)$
Now, $P^{'}$ translated 2 units in negative direction $P^{'} = (6 - 2, 3)$ $P^{'} = (4, 3)$
Now, Convert it to the polar coordinate
$∵ r = \sqrt{4^{2} + 3^{2}} = 5$
$x = r \cos θ$
$y = r \sin θ$
$P^{'} = (r \cos θ, r \sin θ)$
$P^{'} = [r \cos (θ - π / 4), r \sin (θ - π / 4)]$
$P^{'} = [\begin{array}{l} 5 (\cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4}) \\ 5 (\sin θ \cos \frac{π}{4} - \cos θ \sin \frac{π}{4}) \end{array}]$
$P^{'} = [5 (\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{\sqrt{2}}), 5 (\frac{3}{5} \times \frac{1}{\sqrt{2}} - \frac{4}{5} \times \frac{1}{\sqrt{2}})]$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$

AP EAMCET-2013

Co-Ordinate system

88218 If the point $P (1, 3)$ undergoes the following transformations successively.
(i) Reflection with respect to line $y = x$
(ii) Translation through 3 units along the positive direction of the $X$ -axis.
(iii) Rotation through and angles of $\frac{π}{6}$ about the origin in the clockwise direction.
Then the final position of the point $P$ is

1

(\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})

2

(\frac{\sqrt{7}}{2}, \frac{5}{\sqrt{2}})

3

(\frac{6 + \sqrt{3}}{2}, \frac{1 - 6 \sqrt{3}}{2})

4

(\frac{6 + \sqrt{3} - 1}{2}, \frac{6 + \sqrt{3}}{2})

Explanation:

(A) : Given,
The coordinate of the point $P (1, 3)$
We know that,
The point reflection, new position $P_{1} = (3, 1)$

Now,
The translation of point $(3, 1)$ by 3 unit along the positive direction of the $x$ -axis
$P_{2} = (3 + 3, 1)$
$P_{2} = (6, 1)$

Now, The point $P_{2}$ rotate by $π / 6$ then the new position.
$∴ r = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$
$x = r \cos θ$
$x = \sqrt{37} \cos (θ - π / 6)$
$y = r \sin θ$
$y = \sqrt{37} \sin (θ - π / 6)$
Now, $x = \sqrt{37} (\cos θ \cos π / 6 + \sin θ \sin π / 6)$
Here, $\tan θ = 1 / 6$
$\cos θ = \frac{6}{\sqrt{37}} = \frac{6}{r}$
$\sin θ = \frac{1}{\sqrt{37}} = \frac{1}{r}$

$∴ x = ($ r. $\frac{6}{r} \cdot \frac{\sqrt{3}}{2} + r \cdot \frac{1}{r} \cdot \frac{1}{2})$
$x = 3 \sqrt{3} + \frac{1}{2} \Rightarrow x = \frac{6 \sqrt{3} + 1}{2}$
$y = \sqrt{37} (\sin θ \cos π / 6 - \cos θ \sin π / 6)$
$y = ($ r. $\frac{1}{r} \cdot \frac{\sqrt{3}}{2} - r \cdot \frac{6}{r} \cdot \frac{1}{2})$
$y = \frac{\sqrt{3} - 6}{2}$
$∴ coordinate (\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})$

AP EAMCET-2014

Co-Ordinate system

88214 Find the transformed equation of $x \cos θ + y$ $\sin θ = p$ , when the axes are rotated through an angle $θ$ .

1

x = p

2

y = p

3

x + y = p

4

x - y = p

Explanation:

(A) : Given,
The transformed equation is,
$x \cos θ + y \sin θ = p$
The axis of rotation $= θ$
We know that,
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Now, in equation (i) -
$x \cos θ + y \sin θ = p$
$(x^{'} \cos θ - y^{'} \sin θ) \cos θ + (x^{'} \sin θ + y^{'} \cos θ) \sin θ = p$
$x^{'} \cos^{2} θ - y^{'} \sin θ \cos θ + x^{'} \sin^{2} θ + y^{'} \sin θ \cos θ = p$
$x^{'} [\cos^{2} θ + \sin^{2} θ] = p$
$x^{'} = p$
$x = p$

AP EAMCET-2020-17.09.2020

Co-Ordinate system

88215 When the origin is shifted to $(2, 3)$ the transformed equation $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0$ , then the original equation of curve is .......

1

x^{2} - 2 y^{2} - 3 x y + 4 x - y + 20 = 0

2

x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0

3

x^{2} - 2 y^{2} - 3 x y - 4 x - y + 20 = 0

4

x^{2} - 2 y^{2} - 3 x y + 4 x - y - 20 = 0

Explanation:

(B) : Given,
The coordinate of shifted point is $(2, 3)$
The equation is $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0 \dots$ (i)
To get the original equation of curve, replace $(x, y)$ by
$(x - 2, y - 3)$
Now, Putting the value of $x$ and $y$ in equation (i)
$(x - 2)^{2} + 3 (x - 2) (y - 3) - 2 (y - 3)^{2} + 17 (x - 2) - 7 (y - 3)$ $- 11 = 0$
$x^{2} + 4 - 4 x + 3 (x y - 3 x - 2 y + 6) - 2 (y^{2} + 9 - 6 y) + 17 x$ $- 34 - 7 y + 21 - 11 = 0$
$x^{2} + 4 - 4 x + 3 x y - 9 x - 6 y + 18 - 2 y^{2} - 18 + 12 y +$ $17 x - 34 - 7 y + 21 - 11 = 0$
$x^{2} + 3 xy - 2 y^{2} + 4 x - y - 20 = 0$
$x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0$

AP EAMCET-2020-18.09.2020

Co-Ordinate system

88216 The polar equation of the line perpendicular to the line $\sin θ - \cos θ = \frac{1}{r}$ and passing through the point $(2, \frac{π}{6})$ is

1

\sin θ + \cos θ = \frac{\sqrt{3} + 1}{r}

2

\sin θ - \cos θ = \frac{\sqrt{3} + 1}{r}

3

\sin θ + \cos θ = \frac{\sqrt{3} - 1}{r}

4

\cos θ - \sin θ = \frac{\sqrt{3}}{r}

Explanation:

(A) : Given,
The polar equation is, $\sin θ - \cos θ = \frac{1}{r}$
$∴ r \sin θ - r \cos θ = 1$
The perpendicular equation passing through
the point $(2, π / 6)$
We know that,
The conversion of polar coordinate into cartesian coordinate
$x = r \cos θ$
$y = r \sin θ$
$∴$ From equation (i)
$y - x = 1$
The slope of equation (ii) is
$m_{1} = 1$
The slope of perpendicular equation $m_{2} = - 1$
Now, The equation of line which passes through point $(2, π / 6)$
$(y - y_{1}) = m_{2} (x - x_{1})$
$(r \sin θ - 2 \sin π / 6) = - 1 (r \cos θ - 2 \cos π / 6)$
$(r \sin θ - 2 \times \frac{1}{2}) = - 1 (r \cos θ - 2 \times \frac{\sqrt{3}}{2})$
$r \sin θ - 1 = - r \cos θ + \sqrt{3}$
$r (\sin θ + \cos θ) = 1 + \sqrt{3}$
$\sin θ + \cos θ = \frac{1 + \sqrt{3}}{r}$

AP EAMCET-2011

Co-Ordinate system

88217 The origin is translated to $(1, 2)$ . The point $(7$ , 5) in the old system undergoes the following transformations successively.
I. Moves to the new point under the given translation of origin.
II. Translated through 2 units along the negative direction of the new $X$ -axis.
III. Rotated through an angle $\frac{π}{4}$ about the origin of new system in the clockwise direction. The final position of the point $(7, 5)$ is

1

(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

2

(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

(C) : Given,
The old coordinate of point $P = (7, 5)$
The coordinate shifted by $(1, 2)$
We know that,
The new coordinate after translation
$P^{'} = (7 - 1, 5 - 2)$
$P^{'} = (6, 3)$
Now, $P^{'}$ translated 2 units in negative direction $P^{'} = (6 - 2, 3)$ $P^{'} = (4, 3)$
Now, Convert it to the polar coordinate
$∵ r = \sqrt{4^{2} + 3^{2}} = 5$
$x = r \cos θ$
$y = r \sin θ$
$P^{'} = (r \cos θ, r \sin θ)$
$P^{'} = [r \cos (θ - π / 4), r \sin (θ - π / 4)]$
$P^{'} = [\begin{array}{l} 5 (\cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4}) \\ 5 (\sin θ \cos \frac{π}{4} - \cos θ \sin \frac{π}{4}) \end{array}]$
$P^{'} = [5 (\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{\sqrt{2}}), 5 (\frac{3}{5} \times \frac{1}{\sqrt{2}} - \frac{4}{5} \times \frac{1}{\sqrt{2}})]$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$

AP EAMCET-2013

Co-Ordinate system

88218 If the point $P (1, 3)$ undergoes the following transformations successively.
(i) Reflection with respect to line $y = x$
(ii) Translation through 3 units along the positive direction of the $X$ -axis.
(iii) Rotation through and angles of $\frac{π}{6}$ about the origin in the clockwise direction.
Then the final position of the point $P$ is

1

(\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})

2

(\frac{\sqrt{7}}{2}, \frac{5}{\sqrt{2}})

3

(\frac{6 + \sqrt{3}}{2}, \frac{1 - 6 \sqrt{3}}{2})

4

(\frac{6 + \sqrt{3} - 1}{2}, \frac{6 + \sqrt{3}}{2})

Explanation:

(A) : Given,
The coordinate of the point $P (1, 3)$
We know that,
The point reflection, new position $P_{1} = (3, 1)$

Now,
The translation of point $(3, 1)$ by 3 unit along the positive direction of the $x$ -axis
$P_{2} = (3 + 3, 1)$
$P_{2} = (6, 1)$

Now, The point $P_{2}$ rotate by $π / 6$ then the new position.
$∴ r = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$
$x = r \cos θ$
$x = \sqrt{37} \cos (θ - π / 6)$
$y = r \sin θ$
$y = \sqrt{37} \sin (θ - π / 6)$
Now, $x = \sqrt{37} (\cos θ \cos π / 6 + \sin θ \sin π / 6)$
Here, $\tan θ = 1 / 6$
$\cos θ = \frac{6}{\sqrt{37}} = \frac{6}{r}$
$\sin θ = \frac{1}{\sqrt{37}} = \frac{1}{r}$

$∴ x = ($ r. $\frac{6}{r} \cdot \frac{\sqrt{3}}{2} + r \cdot \frac{1}{r} \cdot \frac{1}{2})$
$x = 3 \sqrt{3} + \frac{1}{2} \Rightarrow x = \frac{6 \sqrt{3} + 1}{2}$
$y = \sqrt{37} (\sin θ \cos π / 6 - \cos θ \sin π / 6)$
$y = ($ r. $\frac{1}{r} \cdot \frac{\sqrt{3}}{2} - r \cdot \frac{6}{r} \cdot \frac{1}{2})$
$y = \frac{\sqrt{3} - 6}{2}$
$∴ coordinate (\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})$

AP EAMCET-2014

Co-Ordinate system

88214 Find the transformed equation of $x \cos θ + y$ $\sin θ = p$ , when the axes are rotated through an angle $θ$ .

1

x = p

2

y = p

3

x + y = p

4

x - y = p

Explanation:

(A) : Given,
The transformed equation is,
$x \cos θ + y \sin θ = p$
The axis of rotation $= θ$
We know that,
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Now, in equation (i) -
$x \cos θ + y \sin θ = p$
$(x^{'} \cos θ - y^{'} \sin θ) \cos θ + (x^{'} \sin θ + y^{'} \cos θ) \sin θ = p$
$x^{'} \cos^{2} θ - y^{'} \sin θ \cos θ + x^{'} \sin^{2} θ + y^{'} \sin θ \cos θ = p$
$x^{'} [\cos^{2} θ + \sin^{2} θ] = p$
$x^{'} = p$
$x = p$

AP EAMCET-2020-17.09.2020

Co-Ordinate system

88215 When the origin is shifted to $(2, 3)$ the transformed equation $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0$ , then the original equation of curve is .......

1

x^{2} - 2 y^{2} - 3 x y + 4 x - y + 20 = 0

2

x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0

3

x^{2} - 2 y^{2} - 3 x y - 4 x - y + 20 = 0

4

x^{2} - 2 y^{2} - 3 x y + 4 x - y - 20 = 0

Explanation:

(B) : Given,
The coordinate of shifted point is $(2, 3)$
The equation is $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0 \dots$ (i)
To get the original equation of curve, replace $(x, y)$ by
$(x - 2, y - 3)$
Now, Putting the value of $x$ and $y$ in equation (i)
$(x - 2)^{2} + 3 (x - 2) (y - 3) - 2 (y - 3)^{2} + 17 (x - 2) - 7 (y - 3)$ $- 11 = 0$
$x^{2} + 4 - 4 x + 3 (x y - 3 x - 2 y + 6) - 2 (y^{2} + 9 - 6 y) + 17 x$ $- 34 - 7 y + 21 - 11 = 0$
$x^{2} + 4 - 4 x + 3 x y - 9 x - 6 y + 18 - 2 y^{2} - 18 + 12 y +$ $17 x - 34 - 7 y + 21 - 11 = 0$
$x^{2} + 3 xy - 2 y^{2} + 4 x - y - 20 = 0$
$x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0$

AP EAMCET-2020-18.09.2020

Co-Ordinate system

88216 The polar equation of the line perpendicular to the line $\sin θ - \cos θ = \frac{1}{r}$ and passing through the point $(2, \frac{π}{6})$ is

1

\sin θ + \cos θ = \frac{\sqrt{3} + 1}{r}

2

\sin θ - \cos θ = \frac{\sqrt{3} + 1}{r}

3

\sin θ + \cos θ = \frac{\sqrt{3} - 1}{r}

4

\cos θ - \sin θ = \frac{\sqrt{3}}{r}

Explanation:

(A) : Given,
The polar equation is, $\sin θ - \cos θ = \frac{1}{r}$
$∴ r \sin θ - r \cos θ = 1$
The perpendicular equation passing through
the point $(2, π / 6)$
We know that,
The conversion of polar coordinate into cartesian coordinate
$x = r \cos θ$
$y = r \sin θ$
$∴$ From equation (i)
$y - x = 1$
The slope of equation (ii) is
$m_{1} = 1$
The slope of perpendicular equation $m_{2} = - 1$
Now, The equation of line which passes through point $(2, π / 6)$
$(y - y_{1}) = m_{2} (x - x_{1})$
$(r \sin θ - 2 \sin π / 6) = - 1 (r \cos θ - 2 \cos π / 6)$
$(r \sin θ - 2 \times \frac{1}{2}) = - 1 (r \cos θ - 2 \times \frac{\sqrt{3}}{2})$
$r \sin θ - 1 = - r \cos θ + \sqrt{3}$
$r (\sin θ + \cos θ) = 1 + \sqrt{3}$
$\sin θ + \cos θ = \frac{1 + \sqrt{3}}{r}$

AP EAMCET-2011

Co-Ordinate system

88217 The origin is translated to $(1, 2)$ . The point $(7$ , 5) in the old system undergoes the following transformations successively.
I. Moves to the new point under the given translation of origin.
II. Translated through 2 units along the negative direction of the new $X$ -axis.
III. Rotated through an angle $\frac{π}{4}$ about the origin of new system in the clockwise direction. The final position of the point $(7, 5)$ is

1

(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

2

(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

(C) : Given,
The old coordinate of point $P = (7, 5)$
The coordinate shifted by $(1, 2)$
We know that,
The new coordinate after translation
$P^{'} = (7 - 1, 5 - 2)$
$P^{'} = (6, 3)$
Now, $P^{'}$ translated 2 units in negative direction $P^{'} = (6 - 2, 3)$ $P^{'} = (4, 3)$
Now, Convert it to the polar coordinate
$∵ r = \sqrt{4^{2} + 3^{2}} = 5$
$x = r \cos θ$
$y = r \sin θ$
$P^{'} = (r \cos θ, r \sin θ)$
$P^{'} = [r \cos (θ - π / 4), r \sin (θ - π / 4)]$
$P^{'} = [\begin{array}{l} 5 (\cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4}) \\ 5 (\sin θ \cos \frac{π}{4} - \cos θ \sin \frac{π}{4}) \end{array}]$
$P^{'} = [5 (\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{\sqrt{2}}), 5 (\frac{3}{5} \times \frac{1}{\sqrt{2}} - \frac{4}{5} \times \frac{1}{\sqrt{2}})]$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$

AP EAMCET-2013

Co-Ordinate system

88218 If the point $P (1, 3)$ undergoes the following transformations successively.
(i) Reflection with respect to line $y = x$
(ii) Translation through 3 units along the positive direction of the $X$ -axis.
(iii) Rotation through and angles of $\frac{π}{6}$ about the origin in the clockwise direction.
Then the final position of the point $P$ is

1

(\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})

2

(\frac{\sqrt{7}}{2}, \frac{5}{\sqrt{2}})

3

(\frac{6 + \sqrt{3}}{2}, \frac{1 - 6 \sqrt{3}}{2})

4

(\frac{6 + \sqrt{3} - 1}{2}, \frac{6 + \sqrt{3}}{2})

Explanation:

(A) : Given,
The coordinate of the point $P (1, 3)$
We know that,
The point reflection, new position $P_{1} = (3, 1)$

Now,
The translation of point $(3, 1)$ by 3 unit along the positive direction of the $x$ -axis
$P_{2} = (3 + 3, 1)$
$P_{2} = (6, 1)$

Now, The point $P_{2}$ rotate by $π / 6$ then the new position.
$∴ r = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$
$x = r \cos θ$
$x = \sqrt{37} \cos (θ - π / 6)$
$y = r \sin θ$
$y = \sqrt{37} \sin (θ - π / 6)$
Now, $x = \sqrt{37} (\cos θ \cos π / 6 + \sin θ \sin π / 6)$
Here, $\tan θ = 1 / 6$
$\cos θ = \frac{6}{\sqrt{37}} = \frac{6}{r}$
$\sin θ = \frac{1}{\sqrt{37}} = \frac{1}{r}$

$∴ x = ($ r. $\frac{6}{r} \cdot \frac{\sqrt{3}}{2} + r \cdot \frac{1}{r} \cdot \frac{1}{2})$
$x = 3 \sqrt{3} + \frac{1}{2} \Rightarrow x = \frac{6 \sqrt{3} + 1}{2}$
$y = \sqrt{37} (\sin θ \cos π / 6 - \cos θ \sin π / 6)$
$y = ($ r. $\frac{1}{r} \cdot \frac{\sqrt{3}}{2} - r \cdot \frac{6}{r} \cdot \frac{1}{2})$
$y = \frac{\sqrt{3} - 6}{2}$
$∴ coordinate (\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})$

AP EAMCET-2014

Co-Ordinate system

88214 Find the transformed equation of $x \cos θ + y$ $\sin θ = p$ , when the axes are rotated through an angle $θ$ .

1

x = p

2

y = p

3

x + y = p

4

x - y = p

Explanation:

(A) : Given,
The transformed equation is,
$x \cos θ + y \sin θ = p$
The axis of rotation $= θ$
We know that,
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Now, in equation (i) -
$x \cos θ + y \sin θ = p$
$(x^{'} \cos θ - y^{'} \sin θ) \cos θ + (x^{'} \sin θ + y^{'} \cos θ) \sin θ = p$
$x^{'} \cos^{2} θ - y^{'} \sin θ \cos θ + x^{'} \sin^{2} θ + y^{'} \sin θ \cos θ = p$
$x^{'} [\cos^{2} θ + \sin^{2} θ] = p$
$x^{'} = p$
$x = p$

AP EAMCET-2020-17.09.2020

Co-Ordinate system

88215 When the origin is shifted to $(2, 3)$ the transformed equation $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0$ , then the original equation of curve is .......

1

x^{2} - 2 y^{2} - 3 x y + 4 x - y + 20 = 0

2

x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0

3

x^{2} - 2 y^{2} - 3 x y - 4 x - y + 20 = 0

4

x^{2} - 2 y^{2} - 3 x y + 4 x - y - 20 = 0

Explanation:

(B) : Given,
The coordinate of shifted point is $(2, 3)$
The equation is $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0 \dots$ (i)
To get the original equation of curve, replace $(x, y)$ by
$(x - 2, y - 3)$
Now, Putting the value of $x$ and $y$ in equation (i)
$(x - 2)^{2} + 3 (x - 2) (y - 3) - 2 (y - 3)^{2} + 17 (x - 2) - 7 (y - 3)$ $- 11 = 0$
$x^{2} + 4 - 4 x + 3 (x y - 3 x - 2 y + 6) - 2 (y^{2} + 9 - 6 y) + 17 x$ $- 34 - 7 y + 21 - 11 = 0$
$x^{2} + 4 - 4 x + 3 x y - 9 x - 6 y + 18 - 2 y^{2} - 18 + 12 y +$ $17 x - 34 - 7 y + 21 - 11 = 0$
$x^{2} + 3 xy - 2 y^{2} + 4 x - y - 20 = 0$
$x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0$

AP EAMCET-2020-18.09.2020

Co-Ordinate system

88216 The polar equation of the line perpendicular to the line $\sin θ - \cos θ = \frac{1}{r}$ and passing through the point $(2, \frac{π}{6})$ is

1

\sin θ + \cos θ = \frac{\sqrt{3} + 1}{r}

2

\sin θ - \cos θ = \frac{\sqrt{3} + 1}{r}

3

\sin θ + \cos θ = \frac{\sqrt{3} - 1}{r}

4

\cos θ - \sin θ = \frac{\sqrt{3}}{r}

Explanation:

(A) : Given,
The polar equation is, $\sin θ - \cos θ = \frac{1}{r}$
$∴ r \sin θ - r \cos θ = 1$
The perpendicular equation passing through
the point $(2, π / 6)$
We know that,
The conversion of polar coordinate into cartesian coordinate
$x = r \cos θ$
$y = r \sin θ$
$∴$ From equation (i)
$y - x = 1$
The slope of equation (ii) is
$m_{1} = 1$
The slope of perpendicular equation $m_{2} = - 1$
Now, The equation of line which passes through point $(2, π / 6)$
$(y - y_{1}) = m_{2} (x - x_{1})$
$(r \sin θ - 2 \sin π / 6) = - 1 (r \cos θ - 2 \cos π / 6)$
$(r \sin θ - 2 \times \frac{1}{2}) = - 1 (r \cos θ - 2 \times \frac{\sqrt{3}}{2})$
$r \sin θ - 1 = - r \cos θ + \sqrt{3}$
$r (\sin θ + \cos θ) = 1 + \sqrt{3}$
$\sin θ + \cos θ = \frac{1 + \sqrt{3}}{r}$

AP EAMCET-2011

Co-Ordinate system

88217 The origin is translated to $(1, 2)$ . The point $(7$ , 5) in the old system undergoes the following transformations successively.
I. Moves to the new point under the given translation of origin.
II. Translated through 2 units along the negative direction of the new $X$ -axis.
III. Rotated through an angle $\frac{π}{4}$ about the origin of new system in the clockwise direction. The final position of the point $(7, 5)$ is

1

(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

2

(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

(C) : Given,
The old coordinate of point $P = (7, 5)$
The coordinate shifted by $(1, 2)$
We know that,
The new coordinate after translation
$P^{'} = (7 - 1, 5 - 2)$
$P^{'} = (6, 3)$
Now, $P^{'}$ translated 2 units in negative direction $P^{'} = (6 - 2, 3)$ $P^{'} = (4, 3)$
Now, Convert it to the polar coordinate
$∵ r = \sqrt{4^{2} + 3^{2}} = 5$
$x = r \cos θ$
$y = r \sin θ$
$P^{'} = (r \cos θ, r \sin θ)$
$P^{'} = [r \cos (θ - π / 4), r \sin (θ - π / 4)]$
$P^{'} = [\begin{array}{l} 5 (\cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4}) \\ 5 (\sin θ \cos \frac{π}{4} - \cos θ \sin \frac{π}{4}) \end{array}]$
$P^{'} = [5 (\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{\sqrt{2}}), 5 (\frac{3}{5} \times \frac{1}{\sqrt{2}} - \frac{4}{5} \times \frac{1}{\sqrt{2}})]$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$

AP EAMCET-2013

Co-Ordinate system

88218 If the point $P (1, 3)$ undergoes the following transformations successively.
(i) Reflection with respect to line $y = x$
(ii) Translation through 3 units along the positive direction of the $X$ -axis.
(iii) Rotation through and angles of $\frac{π}{6}$ about the origin in the clockwise direction.
Then the final position of the point $P$ is

1

(\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})

2

(\frac{\sqrt{7}}{2}, \frac{5}{\sqrt{2}})

3

(\frac{6 + \sqrt{3}}{2}, \frac{1 - 6 \sqrt{3}}{2})

4

(\frac{6 + \sqrt{3} - 1}{2}, \frac{6 + \sqrt{3}}{2})

Explanation:

(A) : Given,
The coordinate of the point $P (1, 3)$
We know that,
The point reflection, new position $P_{1} = (3, 1)$

Now,
The translation of point $(3, 1)$ by 3 unit along the positive direction of the $x$ -axis
$P_{2} = (3 + 3, 1)$
$P_{2} = (6, 1)$

Now, The point $P_{2}$ rotate by $π / 6$ then the new position.
$∴ r = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$
$x = r \cos θ$
$x = \sqrt{37} \cos (θ - π / 6)$
$y = r \sin θ$
$y = \sqrt{37} \sin (θ - π / 6)$
Now, $x = \sqrt{37} (\cos θ \cos π / 6 + \sin θ \sin π / 6)$
Here, $\tan θ = 1 / 6$
$\cos θ = \frac{6}{\sqrt{37}} = \frac{6}{r}$
$\sin θ = \frac{1}{\sqrt{37}} = \frac{1}{r}$

$∴ x = ($ r. $\frac{6}{r} \cdot \frac{\sqrt{3}}{2} + r \cdot \frac{1}{r} \cdot \frac{1}{2})$
$x = 3 \sqrt{3} + \frac{1}{2} \Rightarrow x = \frac{6 \sqrt{3} + 1}{2}$
$y = \sqrt{37} (\sin θ \cos π / 6 - \cos θ \sin π / 6)$
$y = ($ r. $\frac{1}{r} \cdot \frac{\sqrt{3}}{2} - r \cdot \frac{6}{r} \cdot \frac{1}{2})$
$y = \frac{\sqrt{3} - 6}{2}$
$∴ coordinate (\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})$

AP EAMCET-2014

Co-Ordinate system

88214 Find the transformed equation of $x \cos θ + y$ $\sin θ = p$ , when the axes are rotated through an angle $θ$ .

1

x = p

2

y = p

3

x + y = p

4

x - y = p

Explanation:

(A) : Given,
The transformed equation is,
$x \cos θ + y \sin θ = p$
The axis of rotation $= θ$
We know that,
$x = x^{'} \cos θ - y^{'} \sin θ$
$y = x^{'} \sin θ + y^{'} \cos θ$
Now, in equation (i) -
$x \cos θ + y \sin θ = p$
$(x^{'} \cos θ - y^{'} \sin θ) \cos θ + (x^{'} \sin θ + y^{'} \cos θ) \sin θ = p$
$x^{'} \cos^{2} θ - y^{'} \sin θ \cos θ + x^{'} \sin^{2} θ + y^{'} \sin θ \cos θ = p$
$x^{'} [\cos^{2} θ + \sin^{2} θ] = p$
$x^{'} = p$
$x = p$

AP EAMCET-2020-17.09.2020

Co-Ordinate system

88215 When the origin is shifted to $(2, 3)$ the transformed equation $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0$ , then the original equation of curve is .......

1

x^{2} - 2 y^{2} - 3 x y + 4 x - y + 20 = 0

2

x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0

3

x^{2} - 2 y^{2} - 3 x y - 4 x - y + 20 = 0

4

x^{2} - 2 y^{2} - 3 x y + 4 x - y - 20 = 0

Explanation:

(B) : Given,
The coordinate of shifted point is $(2, 3)$
The equation is $x^{2} + 3 x y - 2 y^{2} + 17 x - 7 y - 11 = 0 \dots$ (i)
To get the original equation of curve, replace $(x, y)$ by
$(x - 2, y - 3)$
Now, Putting the value of $x$ and $y$ in equation (i)
$(x - 2)^{2} + 3 (x - 2) (y - 3) - 2 (y - 3)^{2} + 17 (x - 2) - 7 (y - 3)$ $- 11 = 0$
$x^{2} + 4 - 4 x + 3 (x y - 3 x - 2 y + 6) - 2 (y^{2} + 9 - 6 y) + 17 x$ $- 34 - 7 y + 21 - 11 = 0$
$x^{2} + 4 - 4 x + 3 x y - 9 x - 6 y + 18 - 2 y^{2} - 18 + 12 y +$ $17 x - 34 - 7 y + 21 - 11 = 0$
$x^{2} + 3 xy - 2 y^{2} + 4 x - y - 20 = 0$
$x^{2} - 2 y^{2} + 3 x y + 4 x - y - 20 = 0$

AP EAMCET-2020-18.09.2020

Co-Ordinate system

88216 The polar equation of the line perpendicular to the line $\sin θ - \cos θ = \frac{1}{r}$ and passing through the point $(2, \frac{π}{6})$ is

1

\sin θ + \cos θ = \frac{\sqrt{3} + 1}{r}

2

\sin θ - \cos θ = \frac{\sqrt{3} + 1}{r}

3

\sin θ + \cos θ = \frac{\sqrt{3} - 1}{r}

4

\cos θ - \sin θ = \frac{\sqrt{3}}{r}

Explanation:

(A) : Given,
The polar equation is, $\sin θ - \cos θ = \frac{1}{r}$
$∴ r \sin θ - r \cos θ = 1$
The perpendicular equation passing through
the point $(2, π / 6)$
We know that,
The conversion of polar coordinate into cartesian coordinate
$x = r \cos θ$
$y = r \sin θ$
$∴$ From equation (i)
$y - x = 1$
The slope of equation (ii) is
$m_{1} = 1$
The slope of perpendicular equation $m_{2} = - 1$
Now, The equation of line which passes through point $(2, π / 6)$
$(y - y_{1}) = m_{2} (x - x_{1})$
$(r \sin θ - 2 \sin π / 6) = - 1 (r \cos θ - 2 \cos π / 6)$
$(r \sin θ - 2 \times \frac{1}{2}) = - 1 (r \cos θ - 2 \times \frac{\sqrt{3}}{2})$
$r \sin θ - 1 = - r \cos θ + \sqrt{3}$
$r (\sin θ + \cos θ) = 1 + \sqrt{3}$
$\sin θ + \cos θ = \frac{1 + \sqrt{3}}{r}$

AP EAMCET-2011

Co-Ordinate system

88217 The origin is translated to $(1, 2)$ . The point $(7$ , 5) in the old system undergoes the following transformations successively.
I. Moves to the new point under the given translation of origin.
II. Translated through 2 units along the negative direction of the new $X$ -axis.
III. Rotated through an angle $\frac{π}{4}$ about the origin of new system in the clockwise direction. The final position of the point $(7, 5)$ is

1

(\frac{9}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

2

(\frac{7}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(\frac{5}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

(C) : Given,
The old coordinate of point $P = (7, 5)$
The coordinate shifted by $(1, 2)$
We know that,
The new coordinate after translation
$P^{'} = (7 - 1, 5 - 2)$
$P^{'} = (6, 3)$
Now, $P^{'}$ translated 2 units in negative direction $P^{'} = (6 - 2, 3)$ $P^{'} = (4, 3)$
Now, Convert it to the polar coordinate
$∵ r = \sqrt{4^{2} + 3^{2}} = 5$
$x = r \cos θ$
$y = r \sin θ$
$P^{'} = (r \cos θ, r \sin θ)$
$P^{'} = [r \cos (θ - π / 4), r \sin (θ - π / 4)]$
$P^{'} = [\begin{array}{l} 5 (\cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4}) \\ 5 (\sin θ \cos \frac{π}{4} - \cos θ \sin \frac{π}{4}) \end{array}]$
$P^{'} = [5 (\frac{4}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{1}{\sqrt{2}}), 5 (\frac{3}{5} \times \frac{1}{\sqrt{2}} - \frac{4}{5} \times \frac{1}{\sqrt{2}})]$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$
$P^{'} = (\frac{7}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})$

AP EAMCET-2013

Co-Ordinate system

88218 If the point $P (1, 3)$ undergoes the following transformations successively.
(i) Reflection with respect to line $y = x$
(ii) Translation through 3 units along the positive direction of the $X$ -axis.
(iii) Rotation through and angles of $\frac{π}{6}$ about the origin in the clockwise direction.
Then the final position of the point $P$ is

1

(\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})

2

(\frac{\sqrt{7}}{2}, \frac{5}{\sqrt{2}})

3

(\frac{6 + \sqrt{3}}{2}, \frac{1 - 6 \sqrt{3}}{2})

4

(\frac{6 + \sqrt{3} - 1}{2}, \frac{6 + \sqrt{3}}{2})

Explanation:

(A) : Given,
The coordinate of the point $P (1, 3)$
We know that,
The point reflection, new position $P_{1} = (3, 1)$

Now,
The translation of point $(3, 1)$ by 3 unit along the positive direction of the $x$ -axis
$P_{2} = (3 + 3, 1)$
$P_{2} = (6, 1)$

Now, The point $P_{2}$ rotate by $π / 6$ then the new position.
$∴ r = \sqrt{6^{2} + 1^{2}} = \sqrt{37}$
$x = r \cos θ$
$x = \sqrt{37} \cos (θ - π / 6)$
$y = r \sin θ$
$y = \sqrt{37} \sin (θ - π / 6)$
Now, $x = \sqrt{37} (\cos θ \cos π / 6 + \sin θ \sin π / 6)$
Here, $\tan θ = 1 / 6$
$\cos θ = \frac{6}{\sqrt{37}} = \frac{6}{r}$
$\sin θ = \frac{1}{\sqrt{37}} = \frac{1}{r}$

$∴ x = ($ r. $\frac{6}{r} \cdot \frac{\sqrt{3}}{2} + r \cdot \frac{1}{r} \cdot \frac{1}{2})$
$x = 3 \sqrt{3} + \frac{1}{2} \Rightarrow x = \frac{6 \sqrt{3} + 1}{2}$
$y = \sqrt{37} (\sin θ \cos π / 6 - \cos θ \sin π / 6)$
$y = ($ r. $\frac{1}{r} \cdot \frac{\sqrt{3}}{2} - r \cdot \frac{6}{r} \cdot \frac{1}{2})$
$y = \frac{\sqrt{3} - 6}{2}$
$∴ coordinate (\frac{6 \sqrt{3} + 1}{2}, \frac{\sqrt{3} - 6}{2})$

AP EAMCET-2014

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