QBManager

Co-Ordinate system

88429 A straight rod of length 9 units slides with its ends $A, B$ always on the $X$ and $Y$ -axis respectively. Then the locus of the centroid of $△ O A B$ is :

1

x^{2} + y^{2} = 3

2

x^{2} + y^{2} = 9

3

x^{2} + y^{2} = 1

4

x^{2} + y^{2} = 81

Explanation:

(B) : Let the point be, $A (a, 0)$ and $B (0, b)$
then, $a^{2} + b^{2} = 81$
If centroid $(G) = (\frac{a}{3}, \frac{b}{3})$
$a = 3 x$ and $b = 3 y$
Hence, $9 x^{2} + 9 y^{2} = 81$
$x^{2} + y^{2} = 9$

BITSAT-2020

Co-Ordinate system

88430 The locus of a point of intersection of two lines $x \sqrt{3} - y = k \sqrt{3}$ and $\sqrt{3} k x + k y = \sqrt{3}, k \in R$ , describes

1 an ellipse

2 a hyperbola

3 a pair of lines

4 a parabola

Explanation:

(B) : Given,
$x \sqrt{3} - y = k \sqrt{3}$
$\sqrt{3} k x + k y = \sqrt{3}$
We will equate value of $k$ from eq. (1) and (2)
$\frac{x \sqrt{3} - y}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3} + y}$
$∴ (x \sqrt{3} - y) (x \sqrt{3} + y) = (\sqrt{3}) (\sqrt{3})$
$∴ 3 x^{2} - y^{2} = 3 \Rightarrow \frac{3 x^{2}}{3} - \frac{y^{2}}{3} = \frac{3}{3}$
i.e. $\frac{x^{2}}{1} - \frac{y^{2}}{(\sqrt{3})^{2}} = 1$ , which is an equation of a
hyperbola.

MHT CET-2020

Co-Ordinate system

88431 A variable plane remains at constant distance $p$ from the origin. If it meets the coordinate axes at the points $A, B, C$ then the locus of the Centroid of $△ ABC$ is

1

x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}

2

x^{- 3} + y^{- 3} + z^{- 3} = 9 p^{- 3}

3

x^{2} + y^{2} + z^{2} = 9 p^{2}

4

x^{3} + y^{3} + z^{3} = 9 p^{3}

Explanation:

(A) : Let $A \equiv (a, 0, 0), B \equiv (0, b, 0), C \equiv (0, 0, c)$ , then equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Its distance from the origin,
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$
If $(x, y, z)$ be centroid of $△ A B C$ , then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$
Eliminating a,b,c from (i) and (ii) required Locus is
$x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}$

VITEEE-2016

Co-Ordinate system

88432 The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

1

4 \leq x^{2} + y^{2} \leq 64

2

x^{2} + y^{2} \leq 25

3

x^{2} + y^{2} \geq 25

4

3 \leq x^{2} + y^{2} \geq 9

Explanation:

(A) : Let (x, y) be any point in the set.

$C$ lying on the circle $x^{2} + y^{2} = 25$
and radius 3 from the origin lies between $(OA) = 5 - 3$ $= 2$ and $OB = 5 + 3 = 8$
Hence $4 \leq x^{2} + y^{2} \leq 64$

VITEEE-2013

Co-Ordinate system

88433 The line joining $(5, 0)$ to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio $2 : 3$ at $P$ . If $θ$ varies, then the locus of $P$ is

1 a straight line

2 a pair of straight lines

3 a circle

4 None of the above

Explanation:

(C) : Let coordinates of $P$ be $(h, k)$ , then
$h = \frac{2 (10 \cos θ) + 3 (5)}{2 + 3} = 4 \cos θ + 3$
$and k = \frac{2 (10 \sin θ) + 3 (0)}{2 + 3} = 4 \sin θ$
[Using the internal section formula]
$\frac{h - 3}{4} = \cos θ and \frac{k}{4} = \sin θ$
Squaring and adding both of these equations,
$\frac{(h - 3)^{2}}{16} + \frac{k^{2}}{16} = \cos^{2} θ + \sin^{2} θ$
$\Rightarrow (h - 3)^{2} + k^{2} = 16$
Therefore, locus of point $P$ is
$(x - 3)^{2} + y^{2} = 16 which is a circle.$

VITEEE-2010

Co-Ordinate system

88429 A straight rod of length 9 units slides with its ends $A, B$ always on the $X$ and $Y$ -axis respectively. Then the locus of the centroid of $△ O A B$ is :

1

x^{2} + y^{2} = 3

2

x^{2} + y^{2} = 9

3

x^{2} + y^{2} = 1

4

x^{2} + y^{2} = 81

Explanation:

(B) : Let the point be, $A (a, 0)$ and $B (0, b)$
then, $a^{2} + b^{2} = 81$
If centroid $(G) = (\frac{a}{3}, \frac{b}{3})$
$a = 3 x$ and $b = 3 y$
Hence, $9 x^{2} + 9 y^{2} = 81$
$x^{2} + y^{2} = 9$

BITSAT-2020

Co-Ordinate system

88430 The locus of a point of intersection of two lines $x \sqrt{3} - y = k \sqrt{3}$ and $\sqrt{3} k x + k y = \sqrt{3}, k \in R$ , describes

1 an ellipse

2 a hyperbola

3 a pair of lines

4 a parabola

Explanation:

(B) : Given,
$x \sqrt{3} - y = k \sqrt{3}$
$\sqrt{3} k x + k y = \sqrt{3}$
We will equate value of $k$ from eq. (1) and (2)
$\frac{x \sqrt{3} - y}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3} + y}$
$∴ (x \sqrt{3} - y) (x \sqrt{3} + y) = (\sqrt{3}) (\sqrt{3})$
$∴ 3 x^{2} - y^{2} = 3 \Rightarrow \frac{3 x^{2}}{3} - \frac{y^{2}}{3} = \frac{3}{3}$
i.e. $\frac{x^{2}}{1} - \frac{y^{2}}{(\sqrt{3})^{2}} = 1$ , which is an equation of a
hyperbola.

MHT CET-2020

Co-Ordinate system

88431 A variable plane remains at constant distance $p$ from the origin. If it meets the coordinate axes at the points $A, B, C$ then the locus of the Centroid of $△ ABC$ is

1

x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}

2

x^{- 3} + y^{- 3} + z^{- 3} = 9 p^{- 3}

3

x^{2} + y^{2} + z^{2} = 9 p^{2}

4

x^{3} + y^{3} + z^{3} = 9 p^{3}

Explanation:

(A) : Let $A \equiv (a, 0, 0), B \equiv (0, b, 0), C \equiv (0, 0, c)$ , then equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Its distance from the origin,
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$
If $(x, y, z)$ be centroid of $△ A B C$ , then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$
Eliminating a,b,c from (i) and (ii) required Locus is
$x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}$

VITEEE-2016

Co-Ordinate system

88432 The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

1

4 \leq x^{2} + y^{2} \leq 64

2

x^{2} + y^{2} \leq 25

3

x^{2} + y^{2} \geq 25

4

3 \leq x^{2} + y^{2} \geq 9

Explanation:

(A) : Let (x, y) be any point in the set.

$C$ lying on the circle $x^{2} + y^{2} = 25$
and radius 3 from the origin lies between $(OA) = 5 - 3$ $= 2$ and $OB = 5 + 3 = 8$
Hence $4 \leq x^{2} + y^{2} \leq 64$

VITEEE-2013

Co-Ordinate system

88433 The line joining $(5, 0)$ to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio $2 : 3$ at $P$ . If $θ$ varies, then the locus of $P$ is

1 a straight line

2 a pair of straight lines

3 a circle

4 None of the above

Explanation:

(C) : Let coordinates of $P$ be $(h, k)$ , then
$h = \frac{2 (10 \cos θ) + 3 (5)}{2 + 3} = 4 \cos θ + 3$
$and k = \frac{2 (10 \sin θ) + 3 (0)}{2 + 3} = 4 \sin θ$
[Using the internal section formula]
$\frac{h - 3}{4} = \cos θ and \frac{k}{4} = \sin θ$
Squaring and adding both of these equations,
$\frac{(h - 3)^{2}}{16} + \frac{k^{2}}{16} = \cos^{2} θ + \sin^{2} θ$
$\Rightarrow (h - 3)^{2} + k^{2} = 16$
Therefore, locus of point $P$ is
$(x - 3)^{2} + y^{2} = 16 which is a circle.$

VITEEE-2010

Co-Ordinate system

88429 A straight rod of length 9 units slides with its ends $A, B$ always on the $X$ and $Y$ -axis respectively. Then the locus of the centroid of $△ O A B$ is :

1

x^{2} + y^{2} = 3

2

x^{2} + y^{2} = 9

3

x^{2} + y^{2} = 1

4

x^{2} + y^{2} = 81

Explanation:

(B) : Let the point be, $A (a, 0)$ and $B (0, b)$
then, $a^{2} + b^{2} = 81$
If centroid $(G) = (\frac{a}{3}, \frac{b}{3})$
$a = 3 x$ and $b = 3 y$
Hence, $9 x^{2} + 9 y^{2} = 81$
$x^{2} + y^{2} = 9$

BITSAT-2020

Co-Ordinate system

88430 The locus of a point of intersection of two lines $x \sqrt{3} - y = k \sqrt{3}$ and $\sqrt{3} k x + k y = \sqrt{3}, k \in R$ , describes

1 an ellipse

2 a hyperbola

3 a pair of lines

4 a parabola

Explanation:

(B) : Given,
$x \sqrt{3} - y = k \sqrt{3}$
$\sqrt{3} k x + k y = \sqrt{3}$
We will equate value of $k$ from eq. (1) and (2)
$\frac{x \sqrt{3} - y}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3} + y}$
$∴ (x \sqrt{3} - y) (x \sqrt{3} + y) = (\sqrt{3}) (\sqrt{3})$
$∴ 3 x^{2} - y^{2} = 3 \Rightarrow \frac{3 x^{2}}{3} - \frac{y^{2}}{3} = \frac{3}{3}$
i.e. $\frac{x^{2}}{1} - \frac{y^{2}}{(\sqrt{3})^{2}} = 1$ , which is an equation of a
hyperbola.

MHT CET-2020

Co-Ordinate system

88431 A variable plane remains at constant distance $p$ from the origin. If it meets the coordinate axes at the points $A, B, C$ then the locus of the Centroid of $△ ABC$ is

1

x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}

2

x^{- 3} + y^{- 3} + z^{- 3} = 9 p^{- 3}

3

x^{2} + y^{2} + z^{2} = 9 p^{2}

4

x^{3} + y^{3} + z^{3} = 9 p^{3}

Explanation:

(A) : Let $A \equiv (a, 0, 0), B \equiv (0, b, 0), C \equiv (0, 0, c)$ , then equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Its distance from the origin,
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$
If $(x, y, z)$ be centroid of $△ A B C$ , then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$
Eliminating a,b,c from (i) and (ii) required Locus is
$x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}$

VITEEE-2016

Co-Ordinate system

88432 The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

1

4 \leq x^{2} + y^{2} \leq 64

2

x^{2} + y^{2} \leq 25

3

x^{2} + y^{2} \geq 25

4

3 \leq x^{2} + y^{2} \geq 9

Explanation:

(A) : Let (x, y) be any point in the set.

$C$ lying on the circle $x^{2} + y^{2} = 25$
and radius 3 from the origin lies between $(OA) = 5 - 3$ $= 2$ and $OB = 5 + 3 = 8$
Hence $4 \leq x^{2} + y^{2} \leq 64$

VITEEE-2013

Co-Ordinate system

88433 The line joining $(5, 0)$ to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio $2 : 3$ at $P$ . If $θ$ varies, then the locus of $P$ is

1 a straight line

2 a pair of straight lines

3 a circle

4 None of the above

Explanation:

(C) : Let coordinates of $P$ be $(h, k)$ , then
$h = \frac{2 (10 \cos θ) + 3 (5)}{2 + 3} = 4 \cos θ + 3$
$and k = \frac{2 (10 \sin θ) + 3 (0)}{2 + 3} = 4 \sin θ$
[Using the internal section formula]
$\frac{h - 3}{4} = \cos θ and \frac{k}{4} = \sin θ$
Squaring and adding both of these equations,
$\frac{(h - 3)^{2}}{16} + \frac{k^{2}}{16} = \cos^{2} θ + \sin^{2} θ$
$\Rightarrow (h - 3)^{2} + k^{2} = 16$
Therefore, locus of point $P$ is
$(x - 3)^{2} + y^{2} = 16 which is a circle.$

VITEEE-2010

Co-Ordinate system

88429 A straight rod of length 9 units slides with its ends $A, B$ always on the $X$ and $Y$ -axis respectively. Then the locus of the centroid of $△ O A B$ is :

1

x^{2} + y^{2} = 3

2

x^{2} + y^{2} = 9

3

x^{2} + y^{2} = 1

4

x^{2} + y^{2} = 81

Explanation:

(B) : Let the point be, $A (a, 0)$ and $B (0, b)$
then, $a^{2} + b^{2} = 81$
If centroid $(G) = (\frac{a}{3}, \frac{b}{3})$
$a = 3 x$ and $b = 3 y$
Hence, $9 x^{2} + 9 y^{2} = 81$
$x^{2} + y^{2} = 9$

BITSAT-2020

Co-Ordinate system

88430 The locus of a point of intersection of two lines $x \sqrt{3} - y = k \sqrt{3}$ and $\sqrt{3} k x + k y = \sqrt{3}, k \in R$ , describes

1 an ellipse

2 a hyperbola

3 a pair of lines

4 a parabola

Explanation:

(B) : Given,
$x \sqrt{3} - y = k \sqrt{3}$
$\sqrt{3} k x + k y = \sqrt{3}$
We will equate value of $k$ from eq. (1) and (2)
$\frac{x \sqrt{3} - y}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3} + y}$
$∴ (x \sqrt{3} - y) (x \sqrt{3} + y) = (\sqrt{3}) (\sqrt{3})$
$∴ 3 x^{2} - y^{2} = 3 \Rightarrow \frac{3 x^{2}}{3} - \frac{y^{2}}{3} = \frac{3}{3}$
i.e. $\frac{x^{2}}{1} - \frac{y^{2}}{(\sqrt{3})^{2}} = 1$ , which is an equation of a
hyperbola.

MHT CET-2020

Co-Ordinate system

88431 A variable plane remains at constant distance $p$ from the origin. If it meets the coordinate axes at the points $A, B, C$ then the locus of the Centroid of $△ ABC$ is

1

x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}

2

x^{- 3} + y^{- 3} + z^{- 3} = 9 p^{- 3}

3

x^{2} + y^{2} + z^{2} = 9 p^{2}

4

x^{3} + y^{3} + z^{3} = 9 p^{3}

Explanation:

(A) : Let $A \equiv (a, 0, 0), B \equiv (0, b, 0), C \equiv (0, 0, c)$ , then equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Its distance from the origin,
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$
If $(x, y, z)$ be centroid of $△ A B C$ , then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$
Eliminating a,b,c from (i) and (ii) required Locus is
$x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}$

VITEEE-2016

Co-Ordinate system

88432 The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

1

4 \leq x^{2} + y^{2} \leq 64

2

x^{2} + y^{2} \leq 25

3

x^{2} + y^{2} \geq 25

4

3 \leq x^{2} + y^{2} \geq 9

Explanation:

(A) : Let (x, y) be any point in the set.

$C$ lying on the circle $x^{2} + y^{2} = 25$
and radius 3 from the origin lies between $(OA) = 5 - 3$ $= 2$ and $OB = 5 + 3 = 8$
Hence $4 \leq x^{2} + y^{2} \leq 64$

VITEEE-2013

Co-Ordinate system

88433 The line joining $(5, 0)$ to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio $2 : 3$ at $P$ . If $θ$ varies, then the locus of $P$ is

1 a straight line

2 a pair of straight lines

3 a circle

4 None of the above

Explanation:

(C) : Let coordinates of $P$ be $(h, k)$ , then
$h = \frac{2 (10 \cos θ) + 3 (5)}{2 + 3} = 4 \cos θ + 3$
$and k = \frac{2 (10 \sin θ) + 3 (0)}{2 + 3} = 4 \sin θ$
[Using the internal section formula]
$\frac{h - 3}{4} = \cos θ and \frac{k}{4} = \sin θ$
Squaring and adding both of these equations,
$\frac{(h - 3)^{2}}{16} + \frac{k^{2}}{16} = \cos^{2} θ + \sin^{2} θ$
$\Rightarrow (h - 3)^{2} + k^{2} = 16$
Therefore, locus of point $P$ is
$(x - 3)^{2} + y^{2} = 16 which is a circle.$

VITEEE-2010

Co-Ordinate system

88429 A straight rod of length 9 units slides with its ends $A, B$ always on the $X$ and $Y$ -axis respectively. Then the locus of the centroid of $△ O A B$ is :

1

x^{2} + y^{2} = 3

2

x^{2} + y^{2} = 9

3

x^{2} + y^{2} = 1

4

x^{2} + y^{2} = 81

Explanation:

(B) : Let the point be, $A (a, 0)$ and $B (0, b)$
then, $a^{2} + b^{2} = 81$
If centroid $(G) = (\frac{a}{3}, \frac{b}{3})$
$a = 3 x$ and $b = 3 y$
Hence, $9 x^{2} + 9 y^{2} = 81$
$x^{2} + y^{2} = 9$

BITSAT-2020

Co-Ordinate system

88430 The locus of a point of intersection of two lines $x \sqrt{3} - y = k \sqrt{3}$ and $\sqrt{3} k x + k y = \sqrt{3}, k \in R$ , describes

1 an ellipse

2 a hyperbola

3 a pair of lines

4 a parabola

Explanation:

(B) : Given,
$x \sqrt{3} - y = k \sqrt{3}$
$\sqrt{3} k x + k y = \sqrt{3}$
We will equate value of $k$ from eq. (1) and (2)
$\frac{x \sqrt{3} - y}{\sqrt{3}} = \frac{\sqrt{3}}{x \sqrt{3} + y}$
$∴ (x \sqrt{3} - y) (x \sqrt{3} + y) = (\sqrt{3}) (\sqrt{3})$
$∴ 3 x^{2} - y^{2} = 3 \Rightarrow \frac{3 x^{2}}{3} - \frac{y^{2}}{3} = \frac{3}{3}$
i.e. $\frac{x^{2}}{1} - \frac{y^{2}}{(\sqrt{3})^{2}} = 1$ , which is an equation of a
hyperbola.

MHT CET-2020

Co-Ordinate system

88431 A variable plane remains at constant distance $p$ from the origin. If it meets the coordinate axes at the points $A, B, C$ then the locus of the Centroid of $△ ABC$ is

1

x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}

2

x^{- 3} + y^{- 3} + z^{- 3} = 9 p^{- 3}

3

x^{2} + y^{2} + z^{2} = 9 p^{2}

4

x^{3} + y^{3} + z^{3} = 9 p^{3}

Explanation:

(A) : Let $A \equiv (a, 0, 0), B \equiv (0, b, 0), C \equiv (0, 0, c)$ , then equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Its distance from the origin,
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$
If $(x, y, z)$ be centroid of $△ A B C$ , then $x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}$
Eliminating a,b,c from (i) and (ii) required Locus is
$x^{- 2} + y^{- 2} + z^{- 2} = 9 p^{- 2}$

VITEEE-2016

Co-Ordinate system

88432 The centres of a set of circles, each of radius 3, lie on the circle $x^{2} + y^{2} = 25$ . The locus of any point in the set is

1

4 \leq x^{2} + y^{2} \leq 64

2

x^{2} + y^{2} \leq 25

3

x^{2} + y^{2} \geq 25

4

3 \leq x^{2} + y^{2} \geq 9

Explanation:

(A) : Let (x, y) be any point in the set.

$C$ lying on the circle $x^{2} + y^{2} = 25$
and radius 3 from the origin lies between $(OA) = 5 - 3$ $= 2$ and $OB = 5 + 3 = 8$
Hence $4 \leq x^{2} + y^{2} \leq 64$

VITEEE-2013

Co-Ordinate system

88433 The line joining $(5, 0)$ to $(10 \cos θ, 10 \sin θ)$ is divided internally in the ratio $2 : 3$ at $P$ . If $θ$ varies, then the locus of $P$ is

1 a straight line

2 a pair of straight lines

3 a circle

4 None of the above

Explanation:

(C) : Let coordinates of $P$ be $(h, k)$ , then
$h = \frac{2 (10 \cos θ) + 3 (5)}{2 + 3} = 4 \cos θ + 3$
$and k = \frac{2 (10 \sin θ) + 3 (0)}{2 + 3} = 4 \sin θ$
[Using the internal section formula]
$\frac{h - 3}{4} = \cos θ and \frac{k}{4} = \sin θ$
Squaring and adding both of these equations,
$\frac{(h - 3)^{2}}{16} + \frac{k^{2}}{16} = \cos^{2} θ + \sin^{2} θ$
$\Rightarrow (h - 3)^{2} + k^{2} = 16$
Therefore, locus of point $P$ is
$(x - 3)^{2} + y^{2} = 16 which is a circle.$

VITEEE-2010