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Co-Ordinate system

88410 $(- \sqrt{2}, \sqrt{2})$ are Cartesian co-ordinates of the point, then its polar co-ordinates are

1

(4, \frac{5 π}{4})

2

(3, \frac{7 π}{4})

3

(1, \frac{4 π}{3})

4

(2, \frac{3 π}{4})

Explanation:

(D) : Given,
The Cartesian co-ordinates of the point is $(- \sqrt{2}, \sqrt{2})$
We know that,
$r \cos θ = - \sqrt{2}$
$r \sin θ = \sqrt{2}$
Squaring the both equation and adding them,
$r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 + 2$
$r^{2} = 4$
$r = \pm 2$
$\tan θ = (\frac{\sqrt{2}}{- \sqrt{2}})$
$\tan θ = - 1$
$θ = 3 π / 4$
$∴$ Required polar co-ordinate $(r, θ) = (2, \frac{3 π}{4})$

MHT CET-2019

Co-Ordinate system

88411 The polar coordinates of $P$ are $(2, \frac{π}{6})$ . If $Q$ is the image of $P$ about the $X$ -axis, then the polar coordinates of $Q$ are

1

(2, \frac{π}{6})

2

(2, \frac{11 π}{6})

3

(2, \frac{5 π}{6})

4

(2, \frac{π}{3})

Explanation:

(B) : Given,
$P \equiv (2, \frac{π}{6})$
The Cartesian coordinates of $P$ are
$x = 2 \cos \frac{π}{6} = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3}$ and
$y = 2 \sin \frac{π}{6} = 2 (\frac{1}{2}) = 1 \Rightarrow p = (\sqrt{3}, 1)$
As $Q$ is the image of $P$ about the $X$ -axis,
$∴ Q \equiv (\sqrt{3}, - 1)$ i.e. $Q$ lies in $4^{th}$ quadrant
Now $r = \sqrt{3 + 1} = 2$ and
$θ = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = - \tan^{- 1} (\frac{1}{\sqrt{3}}) = 2 π - \frac{π}{6} = \frac{11 π}{6}$
Hence, the polar coordinates of $Q$ are $(2, \frac{11 π}{6})$

MHT CET-2019

Co-Ordinate system

88412 If the point $A (5, k), B (- 3, 1)$ and $C (- 7, - 2)$ are collinear, then $k =$

1

\frac{1}{7}

2 7

3 -7

4

\frac{- 1}{7}

Explanation:

(B) :
Slope of $A B =$ slope of $B C$
$\frac{1 - k}{- 3 - 5} = \frac{- 2 - 1}{- 7 + 3} \Rightarrow \frac{1 - k}{- 8} = \frac{- 3}{- 4}$
$\frac{1 - k}{2} = - 3 \Rightarrow k = 7$

MHT CET-2020

Co-Ordinate system

88413 The Cartesian co-ordinates of the point whose polar co-ordinates are $(\frac{1}{2}, 120^{\circ})$ are

1

(\frac{- 1}{4}, \frac{- \sqrt{3}}{4})

2

(\frac{1}{4}, \frac{\sqrt{3}}{4})

3

(\frac{- 1}{4}, \frac{\sqrt{3}}{4})

4

(\frac{1}{4}, \frac{- \sqrt{3}}{4})

Explanation:

(C) : Given,
$P (r, θ) = (\frac{1}{2}, 120^{\circ}) \Rightarrow r = \frac{1}{2}, θ = 120^{\circ}$
We have,
$x = r \cos θ = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} (- \frac{1}{2}) \Rightarrow x = - \frac{1}{4}$
And $y = r \sin θ = \frac{1}{2} \sin 120^{\circ}$
And $= \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$
$∴$ Required point is $P (- \frac{1}{4}, \frac{\sqrt{3}}{4})$

MHT CET-2020

Co-Ordinate system

88414 A normal to the curve $2 x^{2} - y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3 y = 4$ . Then the point $(x_{1}, y_{1})$ is

1

(3, 3)

2

(- 4, - 2)

3

(2, 3)

4

(3, 2)

Explanation:

(D) : Given,
$2 x^{2} - y^{2} = 14$
Differentiating (i) w.r.t $x$ , we get-
$4 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{2 x}{y}$
Let $m, =$ slope of normal at $(x, y) =, - \frac{y,}{2 x,}$
Also, slope of line $x + 3 y = 4$ is $m_{2} = - \frac{1}{3}$
Since the line and normal to the curve are parallel.
$m_{1} m_{2}$
$\Rightarrow - \frac{y_{1}}{2 x_{1}} = - \frac{1}{3} \Rightarrow 2 x_{1} = 3 y_{1}$
$\Rightarrow y_{1} = \frac{2}{3} x_{1}$
Since $(x_{1}, y_{1})$ lies on (i)
$∴ 2 x_{1}^{2} - y_{1}^{2} = 14 \Rightarrow 2 x_{1}^{2} - \frac{4}{9} x_{1}^{2} = 14$
$14 x_{1}^{2} = 126$
$x_{1} = \pm 3$
$∴ y_{1} = 20 r - 2$
So, point $(x_{1}, y_{1})$ is $(3, 2)$

J&K CET-2016

Co-Ordinate system

88410 $(- \sqrt{2}, \sqrt{2})$ are Cartesian co-ordinates of the point, then its polar co-ordinates are

1

(4, \frac{5 π}{4})

2

(3, \frac{7 π}{4})

3

(1, \frac{4 π}{3})

4

(2, \frac{3 π}{4})

Explanation:

(D) : Given,
The Cartesian co-ordinates of the point is $(- \sqrt{2}, \sqrt{2})$
We know that,
$r \cos θ = - \sqrt{2}$
$r \sin θ = \sqrt{2}$
Squaring the both equation and adding them,
$r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 + 2$
$r^{2} = 4$
$r = \pm 2$
$\tan θ = (\frac{\sqrt{2}}{- \sqrt{2}})$
$\tan θ = - 1$
$θ = 3 π / 4$
$∴$ Required polar co-ordinate $(r, θ) = (2, \frac{3 π}{4})$

MHT CET-2019

Co-Ordinate system

88411 The polar coordinates of $P$ are $(2, \frac{π}{6})$ . If $Q$ is the image of $P$ about the $X$ -axis, then the polar coordinates of $Q$ are

1

(2, \frac{π}{6})

2

(2, \frac{11 π}{6})

3

(2, \frac{5 π}{6})

4

(2, \frac{π}{3})

Explanation:

(B) : Given,
$P \equiv (2, \frac{π}{6})$
The Cartesian coordinates of $P$ are
$x = 2 \cos \frac{π}{6} = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3}$ and
$y = 2 \sin \frac{π}{6} = 2 (\frac{1}{2}) = 1 \Rightarrow p = (\sqrt{3}, 1)$
As $Q$ is the image of $P$ about the $X$ -axis,
$∴ Q \equiv (\sqrt{3}, - 1)$ i.e. $Q$ lies in $4^{th}$ quadrant
Now $r = \sqrt{3 + 1} = 2$ and
$θ = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = - \tan^{- 1} (\frac{1}{\sqrt{3}}) = 2 π - \frac{π}{6} = \frac{11 π}{6}$
Hence, the polar coordinates of $Q$ are $(2, \frac{11 π}{6})$

MHT CET-2019

Co-Ordinate system

88412 If the point $A (5, k), B (- 3, 1)$ and $C (- 7, - 2)$ are collinear, then $k =$

1

\frac{1}{7}

2 7

3 -7

4

\frac{- 1}{7}

Explanation:

(B) :
Slope of $A B =$ slope of $B C$
$\frac{1 - k}{- 3 - 5} = \frac{- 2 - 1}{- 7 + 3} \Rightarrow \frac{1 - k}{- 8} = \frac{- 3}{- 4}$
$\frac{1 - k}{2} = - 3 \Rightarrow k = 7$

MHT CET-2020

Co-Ordinate system

88413 The Cartesian co-ordinates of the point whose polar co-ordinates are $(\frac{1}{2}, 120^{\circ})$ are

1

(\frac{- 1}{4}, \frac{- \sqrt{3}}{4})

2

(\frac{1}{4}, \frac{\sqrt{3}}{4})

3

(\frac{- 1}{4}, \frac{\sqrt{3}}{4})

4

(\frac{1}{4}, \frac{- \sqrt{3}}{4})

Explanation:

(C) : Given,
$P (r, θ) = (\frac{1}{2}, 120^{\circ}) \Rightarrow r = \frac{1}{2}, θ = 120^{\circ}$
We have,
$x = r \cos θ = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} (- \frac{1}{2}) \Rightarrow x = - \frac{1}{4}$
And $y = r \sin θ = \frac{1}{2} \sin 120^{\circ}$
And $= \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$
$∴$ Required point is $P (- \frac{1}{4}, \frac{\sqrt{3}}{4})$

MHT CET-2020

Co-Ordinate system

88414 A normal to the curve $2 x^{2} - y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3 y = 4$ . Then the point $(x_{1}, y_{1})$ is

1

(3, 3)

2

(- 4, - 2)

3

(2, 3)

4

(3, 2)

Explanation:

(D) : Given,
$2 x^{2} - y^{2} = 14$
Differentiating (i) w.r.t $x$ , we get-
$4 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{2 x}{y}$
Let $m, =$ slope of normal at $(x, y) =, - \frac{y,}{2 x,}$
Also, slope of line $x + 3 y = 4$ is $m_{2} = - \frac{1}{3}$
Since the line and normal to the curve are parallel.
$m_{1} m_{2}$
$\Rightarrow - \frac{y_{1}}{2 x_{1}} = - \frac{1}{3} \Rightarrow 2 x_{1} = 3 y_{1}$
$\Rightarrow y_{1} = \frac{2}{3} x_{1}$
Since $(x_{1}, y_{1})$ lies on (i)
$∴ 2 x_{1}^{2} - y_{1}^{2} = 14 \Rightarrow 2 x_{1}^{2} - \frac{4}{9} x_{1}^{2} = 14$
$14 x_{1}^{2} = 126$
$x_{1} = \pm 3$
$∴ y_{1} = 20 r - 2$
So, point $(x_{1}, y_{1})$ is $(3, 2)$

J&K CET-2016

Co-Ordinate system

88410 $(- \sqrt{2}, \sqrt{2})$ are Cartesian co-ordinates of the point, then its polar co-ordinates are

1

(4, \frac{5 π}{4})

2

(3, \frac{7 π}{4})

3

(1, \frac{4 π}{3})

4

(2, \frac{3 π}{4})

Explanation:

(D) : Given,
The Cartesian co-ordinates of the point is $(- \sqrt{2}, \sqrt{2})$
We know that,
$r \cos θ = - \sqrt{2}$
$r \sin θ = \sqrt{2}$
Squaring the both equation and adding them,
$r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 + 2$
$r^{2} = 4$
$r = \pm 2$
$\tan θ = (\frac{\sqrt{2}}{- \sqrt{2}})$
$\tan θ = - 1$
$θ = 3 π / 4$
$∴$ Required polar co-ordinate $(r, θ) = (2, \frac{3 π}{4})$

MHT CET-2019

Co-Ordinate system

88411 The polar coordinates of $P$ are $(2, \frac{π}{6})$ . If $Q$ is the image of $P$ about the $X$ -axis, then the polar coordinates of $Q$ are

1

(2, \frac{π}{6})

2

(2, \frac{11 π}{6})

3

(2, \frac{5 π}{6})

4

(2, \frac{π}{3})

Explanation:

(B) : Given,
$P \equiv (2, \frac{π}{6})$
The Cartesian coordinates of $P$ are
$x = 2 \cos \frac{π}{6} = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3}$ and
$y = 2 \sin \frac{π}{6} = 2 (\frac{1}{2}) = 1 \Rightarrow p = (\sqrt{3}, 1)$
As $Q$ is the image of $P$ about the $X$ -axis,
$∴ Q \equiv (\sqrt{3}, - 1)$ i.e. $Q$ lies in $4^{th}$ quadrant
Now $r = \sqrt{3 + 1} = 2$ and
$θ = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = - \tan^{- 1} (\frac{1}{\sqrt{3}}) = 2 π - \frac{π}{6} = \frac{11 π}{6}$
Hence, the polar coordinates of $Q$ are $(2, \frac{11 π}{6})$

MHT CET-2019

Co-Ordinate system

88412 If the point $A (5, k), B (- 3, 1)$ and $C (- 7, - 2)$ are collinear, then $k =$

1

\frac{1}{7}

2 7

3 -7

4

\frac{- 1}{7}

Explanation:

(B) :
Slope of $A B =$ slope of $B C$
$\frac{1 - k}{- 3 - 5} = \frac{- 2 - 1}{- 7 + 3} \Rightarrow \frac{1 - k}{- 8} = \frac{- 3}{- 4}$
$\frac{1 - k}{2} = - 3 \Rightarrow k = 7$

MHT CET-2020

Co-Ordinate system

88413 The Cartesian co-ordinates of the point whose polar co-ordinates are $(\frac{1}{2}, 120^{\circ})$ are

1

(\frac{- 1}{4}, \frac{- \sqrt{3}}{4})

2

(\frac{1}{4}, \frac{\sqrt{3}}{4})

3

(\frac{- 1}{4}, \frac{\sqrt{3}}{4})

4

(\frac{1}{4}, \frac{- \sqrt{3}}{4})

Explanation:

(C) : Given,
$P (r, θ) = (\frac{1}{2}, 120^{\circ}) \Rightarrow r = \frac{1}{2}, θ = 120^{\circ}$
We have,
$x = r \cos θ = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} (- \frac{1}{2}) \Rightarrow x = - \frac{1}{4}$
And $y = r \sin θ = \frac{1}{2} \sin 120^{\circ}$
And $= \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$
$∴$ Required point is $P (- \frac{1}{4}, \frac{\sqrt{3}}{4})$

MHT CET-2020

Co-Ordinate system

88414 A normal to the curve $2 x^{2} - y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3 y = 4$ . Then the point $(x_{1}, y_{1})$ is

1

(3, 3)

2

(- 4, - 2)

3

(2, 3)

4

(3, 2)

Explanation:

(D) : Given,
$2 x^{2} - y^{2} = 14$
Differentiating (i) w.r.t $x$ , we get-
$4 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{2 x}{y}$
Let $m, =$ slope of normal at $(x, y) =, - \frac{y,}{2 x,}$
Also, slope of line $x + 3 y = 4$ is $m_{2} = - \frac{1}{3}$
Since the line and normal to the curve are parallel.
$m_{1} m_{2}$
$\Rightarrow - \frac{y_{1}}{2 x_{1}} = - \frac{1}{3} \Rightarrow 2 x_{1} = 3 y_{1}$
$\Rightarrow y_{1} = \frac{2}{3} x_{1}$
Since $(x_{1}, y_{1})$ lies on (i)
$∴ 2 x_{1}^{2} - y_{1}^{2} = 14 \Rightarrow 2 x_{1}^{2} - \frac{4}{9} x_{1}^{2} = 14$
$14 x_{1}^{2} = 126$
$x_{1} = \pm 3$
$∴ y_{1} = 20 r - 2$
So, point $(x_{1}, y_{1})$ is $(3, 2)$

J&K CET-2016

Co-Ordinate system

88410 $(- \sqrt{2}, \sqrt{2})$ are Cartesian co-ordinates of the point, then its polar co-ordinates are

1

(4, \frac{5 π}{4})

2

(3, \frac{7 π}{4})

3

(1, \frac{4 π}{3})

4

(2, \frac{3 π}{4})

Explanation:

(D) : Given,
The Cartesian co-ordinates of the point is $(- \sqrt{2}, \sqrt{2})$
We know that,
$r \cos θ = - \sqrt{2}$
$r \sin θ = \sqrt{2}$
Squaring the both equation and adding them,
$r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 + 2$
$r^{2} = 4$
$r = \pm 2$
$\tan θ = (\frac{\sqrt{2}}{- \sqrt{2}})$
$\tan θ = - 1$
$θ = 3 π / 4$
$∴$ Required polar co-ordinate $(r, θ) = (2, \frac{3 π}{4})$

MHT CET-2019

Co-Ordinate system

88411 The polar coordinates of $P$ are $(2, \frac{π}{6})$ . If $Q$ is the image of $P$ about the $X$ -axis, then the polar coordinates of $Q$ are

1

(2, \frac{π}{6})

2

(2, \frac{11 π}{6})

3

(2, \frac{5 π}{6})

4

(2, \frac{π}{3})

Explanation:

(B) : Given,
$P \equiv (2, \frac{π}{6})$
The Cartesian coordinates of $P$ are
$x = 2 \cos \frac{π}{6} = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3}$ and
$y = 2 \sin \frac{π}{6} = 2 (\frac{1}{2}) = 1 \Rightarrow p = (\sqrt{3}, 1)$
As $Q$ is the image of $P$ about the $X$ -axis,
$∴ Q \equiv (\sqrt{3}, - 1)$ i.e. $Q$ lies in $4^{th}$ quadrant
Now $r = \sqrt{3 + 1} = 2$ and
$θ = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = - \tan^{- 1} (\frac{1}{\sqrt{3}}) = 2 π - \frac{π}{6} = \frac{11 π}{6}$
Hence, the polar coordinates of $Q$ are $(2, \frac{11 π}{6})$

MHT CET-2019

Co-Ordinate system

88412 If the point $A (5, k), B (- 3, 1)$ and $C (- 7, - 2)$ are collinear, then $k =$

1

\frac{1}{7}

2 7

3 -7

4

\frac{- 1}{7}

Explanation:

(B) :
Slope of $A B =$ slope of $B C$
$\frac{1 - k}{- 3 - 5} = \frac{- 2 - 1}{- 7 + 3} \Rightarrow \frac{1 - k}{- 8} = \frac{- 3}{- 4}$
$\frac{1 - k}{2} = - 3 \Rightarrow k = 7$

MHT CET-2020

Co-Ordinate system

88413 The Cartesian co-ordinates of the point whose polar co-ordinates are $(\frac{1}{2}, 120^{\circ})$ are

1

(\frac{- 1}{4}, \frac{- \sqrt{3}}{4})

2

(\frac{1}{4}, \frac{\sqrt{3}}{4})

3

(\frac{- 1}{4}, \frac{\sqrt{3}}{4})

4

(\frac{1}{4}, \frac{- \sqrt{3}}{4})

Explanation:

(C) : Given,
$P (r, θ) = (\frac{1}{2}, 120^{\circ}) \Rightarrow r = \frac{1}{2}, θ = 120^{\circ}$
We have,
$x = r \cos θ = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} (- \frac{1}{2}) \Rightarrow x = - \frac{1}{4}$
And $y = r \sin θ = \frac{1}{2} \sin 120^{\circ}$
And $= \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$
$∴$ Required point is $P (- \frac{1}{4}, \frac{\sqrt{3}}{4})$

MHT CET-2020

Co-Ordinate system

88414 A normal to the curve $2 x^{2} - y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3 y = 4$ . Then the point $(x_{1}, y_{1})$ is

1

(3, 3)

2

(- 4, - 2)

3

(2, 3)

4

(3, 2)

Explanation:

(D) : Given,
$2 x^{2} - y^{2} = 14$
Differentiating (i) w.r.t $x$ , we get-
$4 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{2 x}{y}$
Let $m, =$ slope of normal at $(x, y) =, - \frac{y,}{2 x,}$
Also, slope of line $x + 3 y = 4$ is $m_{2} = - \frac{1}{3}$
Since the line and normal to the curve are parallel.
$m_{1} m_{2}$
$\Rightarrow - \frac{y_{1}}{2 x_{1}} = - \frac{1}{3} \Rightarrow 2 x_{1} = 3 y_{1}$
$\Rightarrow y_{1} = \frac{2}{3} x_{1}$
Since $(x_{1}, y_{1})$ lies on (i)
$∴ 2 x_{1}^{2} - y_{1}^{2} = 14 \Rightarrow 2 x_{1}^{2} - \frac{4}{9} x_{1}^{2} = 14$
$14 x_{1}^{2} = 126$
$x_{1} = \pm 3$
$∴ y_{1} = 20 r - 2$
So, point $(x_{1}, y_{1})$ is $(3, 2)$

J&K CET-2016

Co-Ordinate system

88410 $(- \sqrt{2}, \sqrt{2})$ are Cartesian co-ordinates of the point, then its polar co-ordinates are

1

(4, \frac{5 π}{4})

2

(3, \frac{7 π}{4})

3

(1, \frac{4 π}{3})

4

(2, \frac{3 π}{4})

Explanation:

(D) : Given,
The Cartesian co-ordinates of the point is $(- \sqrt{2}, \sqrt{2})$
We know that,
$r \cos θ = - \sqrt{2}$
$r \sin θ = \sqrt{2}$
Squaring the both equation and adding them,
$r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = 2 + 2$
$r^{2} = 4$
$r = \pm 2$
$\tan θ = (\frac{\sqrt{2}}{- \sqrt{2}})$
$\tan θ = - 1$
$θ = 3 π / 4$
$∴$ Required polar co-ordinate $(r, θ) = (2, \frac{3 π}{4})$

MHT CET-2019

Co-Ordinate system

88411 The polar coordinates of $P$ are $(2, \frac{π}{6})$ . If $Q$ is the image of $P$ about the $X$ -axis, then the polar coordinates of $Q$ are

1

(2, \frac{π}{6})

2

(2, \frac{11 π}{6})

3

(2, \frac{5 π}{6})

4

(2, \frac{π}{3})

Explanation:

(B) : Given,
$P \equiv (2, \frac{π}{6})$
The Cartesian coordinates of $P$ are
$x = 2 \cos \frac{π}{6} = 2 (\frac{\sqrt{3}}{2}) = \sqrt{3}$ and
$y = 2 \sin \frac{π}{6} = 2 (\frac{1}{2}) = 1 \Rightarrow p = (\sqrt{3}, 1)$
As $Q$ is the image of $P$ about the $X$ -axis,
$∴ Q \equiv (\sqrt{3}, - 1)$ i.e. $Q$ lies in $4^{th}$ quadrant
Now $r = \sqrt{3 + 1} = 2$ and
$θ = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = - \tan^{- 1} (\frac{1}{\sqrt{3}}) = 2 π - \frac{π}{6} = \frac{11 π}{6}$
Hence, the polar coordinates of $Q$ are $(2, \frac{11 π}{6})$

MHT CET-2019

Co-Ordinate system

88412 If the point $A (5, k), B (- 3, 1)$ and $C (- 7, - 2)$ are collinear, then $k =$

1

\frac{1}{7}

2 7

3 -7

4

\frac{- 1}{7}

Explanation:

(B) :
Slope of $A B =$ slope of $B C$
$\frac{1 - k}{- 3 - 5} = \frac{- 2 - 1}{- 7 + 3} \Rightarrow \frac{1 - k}{- 8} = \frac{- 3}{- 4}$
$\frac{1 - k}{2} = - 3 \Rightarrow k = 7$

MHT CET-2020

Co-Ordinate system

88413 The Cartesian co-ordinates of the point whose polar co-ordinates are $(\frac{1}{2}, 120^{\circ})$ are

1

(\frac{- 1}{4}, \frac{- \sqrt{3}}{4})

2

(\frac{1}{4}, \frac{\sqrt{3}}{4})

3

(\frac{- 1}{4}, \frac{\sqrt{3}}{4})

4

(\frac{1}{4}, \frac{- \sqrt{3}}{4})

Explanation:

(C) : Given,
$P (r, θ) = (\frac{1}{2}, 120^{\circ}) \Rightarrow r = \frac{1}{2}, θ = 120^{\circ}$
We have,
$x = r \cos θ = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} (- \frac{1}{2}) \Rightarrow x = - \frac{1}{4}$
And $y = r \sin θ = \frac{1}{2} \sin 120^{\circ}$
And $= \frac{1}{2} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$
$∴$ Required point is $P (- \frac{1}{4}, \frac{\sqrt{3}}{4})$

MHT CET-2020

Co-Ordinate system

88414 A normal to the curve $2 x^{2} - y^{2} = 14$ at the point $(x_{1}, y_{1})$ is parallel to the straight line $x + 3 y = 4$ . Then the point $(x_{1}, y_{1})$ is

1

(3, 3)

2

(- 4, - 2)

3

(2, 3)

4

(3, 2)

Explanation:

(D) : Given,
$2 x^{2} - y^{2} = 14$
Differentiating (i) w.r.t $x$ , we get-
$4 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{2 x}{y}$
Let $m, =$ slope of normal at $(x, y) =, - \frac{y,}{2 x,}$
Also, slope of line $x + 3 y = 4$ is $m_{2} = - \frac{1}{3}$
Since the line and normal to the curve are parallel.
$m_{1} m_{2}$
$\Rightarrow - \frac{y_{1}}{2 x_{1}} = - \frac{1}{3} \Rightarrow 2 x_{1} = 3 y_{1}$
$\Rightarrow y_{1} = \frac{2}{3} x_{1}$
Since $(x_{1}, y_{1})$ lies on (i)
$∴ 2 x_{1}^{2} - y_{1}^{2} = 14 \Rightarrow 2 x_{1}^{2} - \frac{4}{9} x_{1}^{2} = 14$
$14 x_{1}^{2} = 126$
$x_{1} = \pm 3$
$∴ y_{1} = 20 r - 2$
So, point $(x_{1}, y_{1})$ is $(3, 2)$

J&K CET-2016

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Question Bank Series

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