QBManager

Conic Section

119680 The diameter of $16 x^{2} - 9 y^{2} = 144$ which
conjugate to $x = 2 y$ is

1

y = \frac{32 x}{9}

2

x = \frac{16}{9} y

3

y = \frac{16}{9} x

4 None of these

Explanation:

A The diameters $y = m_{1} x$ and $y = m_{2} x$ are conjugate diameters of the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Here $a^{2} = 9$ and $b^{2} = 16$
Also given $x = 2 y$
Then $m_{1} = \frac{1}{2}$
$∵ \frac{1}{2} \times m_{2} = \frac{16}{9}$
$m_{2} = \frac{32}{9}$
$∴$ Required equation of the diameter is-
$y = \frac{32 x}{9}$

Manipal UGET-201

Conic Section

119690 The polar equation of the circle with centre at $(2, \frac{π}{2})$ and radius 3 units is

1

r^{2} + 4 r \cos θ = 5

2

r^{2} + 4 r \sin θ = 5

3

r^{2} - 4 r \sin θ = 5

4

r^{2} - 4 r \cos θ = 5

Explanation:

C Given, point $(2, \frac{π}{2})$ and $r = 3$ units
We know that, general equation of the circle in polar form is -
$r^{2} - 2 {rr}_{0} \cos (θ - γ) + r_{0}^{2} = a^{2}$
Where $(r_{0}, γ)$ is the centre and $a$ is the radius
So, the equation of the circle with centre $(2, \frac{π}{2})$ and radius of 3 units is given by
$r^{2} - 2 r (2) \cos (θ - \frac{π}{2}) + (2)^{2} = (3)^{2}$
$r^{2} - 4 r \cos {- (\frac{π}{2} - θ)} + 4 = 9$
$r^{2} - 4 r \cos (\frac{π}{2} - θ) = 9 - 4$
$r^{2} - 4 r \sin θ = 9 - 4 {∵ \cos (\frac{π}{2} - θ) = \sin θ$
$r^{2} - 4 r \sin θ = 5$

Assam CEE-2022

Conic Section

119681 The equation of the circle which passes throug the points $(2, 3)$ and $(4, 5)$ and the centre lies 0 the straight line $y - 4 x + 3 = 0$ is

1

x^{2} + y^{2} + 4 x - 10 y + 25 = 0

2

x^{2} + y^{2} - 4 x - 10 y + 16 = 0

3

x^{2} + y^{2} - 4 x - 10 y + 25 = 0

4 None of the above

Explanation:

C Let, $(h, k)$ be the co-ordinate of centre and a be the radius of the circle. Then equation of circle
$(x - h)^{2} + (y - k)^{2} = a^{2}$
$∵ \dots . . (i)$
${CP}^{2} = {CQ}^{2}$
$\Rightarrow h - 2)^{2} + (k - 3)^{2} = (h - 4)^{2} + (k - 5)^{2}$
$\Rightarrow h + k = 7 \dots \dots . (ii)$
Also centre $(h, k)$ lies on the straight line
$∴ y - 4 x + 3 = 0$
$∴ - 4 h + k = - 3 .$
solving equation (2) and (3)
The co-ordinate of centre $= (2, 5)$
$∴ Radius (a) = \sqrt{(2 - 2)^{2} + (5 - 3)^{2}} = CP = CQ$
$= 2$
$∴$ from equation (i) equation of the circle is-
$(x - 2)^{2} + (y - 5)^{2} = 4$
or $x^{2} + y^{2} - 4 x - 10 y + 25 = 0$

Manipal UGET-201

Conic Section

119682 The point $(- 15, 21)$ lies
#[Qdiff: Hard, QCat: Numerical Based, examname: is, By comparing equation (i) with standard equation, $x^{2} + y^{2} = r^{2}$ , $r^{2} = 625$ , $r = 25$ , Distance of the point from the centre is, $d = \sqrt{(0 - (- 15)^{2} + (0 - 21)^{2}}$ , $= \sqrt{225 + 441} = 25.61 > 25$ Thus the point $(- 15, 21)$ lies out-side of circle, 65. Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its centre,

1 Outside the circle

x^{2} + y^{2} = 729

2 Outside the circle

x^{2} + y^{2} = 625

3 Inside the circle

x^{2} + y^{2} = 529

4 Inside the circle

x^{2} + y^{2} = 576

Explanation:

B Given point $(- 15, 21)$
Consider equation $\to x^{2} + y^{2} = 625$
Ans: c
Exp: (C) : Let c be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid-point of the chords of the circle $C$ that subtended an angle of $\frac{2 π}{3}$ at its centre.
In $△ CPB$
$\cos \frac{π}{3} = \frac{CP}{CB}$
$\frac{1}{2} = \frac{\sqrt{h^{2} + k^{2}}}{3}$ $h^{2} + k^{2} = \frac{9}{4}$
original image
Hence, The equation of the locus of the mid-points of the chords of the circle $C$ is $x^{2} + y^{2} = \frac{9}{4}$

is

Conic Section

119680 The diameter of $16 x^{2} - 9 y^{2} = 144$ which
conjugate to $x = 2 y$ is

1

y = \frac{32 x}{9}

2

x = \frac{16}{9} y

3

y = \frac{16}{9} x

4 None of these

Explanation:

A The diameters $y = m_{1} x$ and $y = m_{2} x$ are conjugate diameters of the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Here $a^{2} = 9$ and $b^{2} = 16$
Also given $x = 2 y$
Then $m_{1} = \frac{1}{2}$
$∵ \frac{1}{2} \times m_{2} = \frac{16}{9}$
$m_{2} = \frac{32}{9}$
$∴$ Required equation of the diameter is-
$y = \frac{32 x}{9}$

Manipal UGET-201

Conic Section

119690 The polar equation of the circle with centre at $(2, \frac{π}{2})$ and radius 3 units is

1

r^{2} + 4 r \cos θ = 5

2

r^{2} + 4 r \sin θ = 5

3

r^{2} - 4 r \sin θ = 5

4

r^{2} - 4 r \cos θ = 5

Explanation:

C Given, point $(2, \frac{π}{2})$ and $r = 3$ units
We know that, general equation of the circle in polar form is -
$r^{2} - 2 {rr}_{0} \cos (θ - γ) + r_{0}^{2} = a^{2}$
Where $(r_{0}, γ)$ is the centre and $a$ is the radius
So, the equation of the circle with centre $(2, \frac{π}{2})$ and radius of 3 units is given by
$r^{2} - 2 r (2) \cos (θ - \frac{π}{2}) + (2)^{2} = (3)^{2}$
$r^{2} - 4 r \cos {- (\frac{π}{2} - θ)} + 4 = 9$
$r^{2} - 4 r \cos (\frac{π}{2} - θ) = 9 - 4$
$r^{2} - 4 r \sin θ = 9 - 4 {∵ \cos (\frac{π}{2} - θ) = \sin θ$
$r^{2} - 4 r \sin θ = 5$

Assam CEE-2022

Conic Section

119681 The equation of the circle which passes throug the points $(2, 3)$ and $(4, 5)$ and the centre lies 0 the straight line $y - 4 x + 3 = 0$ is

1

x^{2} + y^{2} + 4 x - 10 y + 25 = 0

2

x^{2} + y^{2} - 4 x - 10 y + 16 = 0

3

x^{2} + y^{2} - 4 x - 10 y + 25 = 0

4 None of the above

Explanation:

C Let, $(h, k)$ be the co-ordinate of centre and a be the radius of the circle. Then equation of circle
$(x - h)^{2} + (y - k)^{2} = a^{2}$
$∵ \dots . . (i)$
${CP}^{2} = {CQ}^{2}$
$\Rightarrow h - 2)^{2} + (k - 3)^{2} = (h - 4)^{2} + (k - 5)^{2}$
$\Rightarrow h + k = 7 \dots \dots . (ii)$
Also centre $(h, k)$ lies on the straight line
$∴ y - 4 x + 3 = 0$
$∴ - 4 h + k = - 3 .$
solving equation (2) and (3)
The co-ordinate of centre $= (2, 5)$
$∴ Radius (a) = \sqrt{(2 - 2)^{2} + (5 - 3)^{2}} = CP = CQ$
$= 2$
$∴$ from equation (i) equation of the circle is-
$(x - 2)^{2} + (y - 5)^{2} = 4$
or $x^{2} + y^{2} - 4 x - 10 y + 25 = 0$

Manipal UGET-201

Conic Section

119682 The point $(- 15, 21)$ lies
#[Qdiff: Hard, QCat: Numerical Based, examname: is, By comparing equation (i) with standard equation, $x^{2} + y^{2} = r^{2}$ , $r^{2} = 625$ , $r = 25$ , Distance of the point from the centre is, $d = \sqrt{(0 - (- 15)^{2} + (0 - 21)^{2}}$ , $= \sqrt{225 + 441} = 25.61 > 25$ Thus the point $(- 15, 21)$ lies out-side of circle, 65. Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its centre,

1 Outside the circle

x^{2} + y^{2} = 729

2 Outside the circle

x^{2} + y^{2} = 625

3 Inside the circle

x^{2} + y^{2} = 529

4 Inside the circle

x^{2} + y^{2} = 576

Explanation:

B Given point $(- 15, 21)$
Consider equation $\to x^{2} + y^{2} = 625$
Ans: c
Exp: (C) : Let c be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid-point of the chords of the circle $C$ that subtended an angle of $\frac{2 π}{3}$ at its centre.
In $△ CPB$
$\cos \frac{π}{3} = \frac{CP}{CB}$
$\frac{1}{2} = \frac{\sqrt{h^{2} + k^{2}}}{3}$ $h^{2} + k^{2} = \frac{9}{4}$
original image
Hence, The equation of the locus of the mid-points of the chords of the circle $C$ is $x^{2} + y^{2} = \frac{9}{4}$

is

Conic Section

119680 The diameter of $16 x^{2} - 9 y^{2} = 144$ which
conjugate to $x = 2 y$ is

1

y = \frac{32 x}{9}

2

x = \frac{16}{9} y

3

y = \frac{16}{9} x

4 None of these

Explanation:

A The diameters $y = m_{1} x$ and $y = m_{2} x$ are conjugate diameters of the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Here $a^{2} = 9$ and $b^{2} = 16$
Also given $x = 2 y$
Then $m_{1} = \frac{1}{2}$
$∵ \frac{1}{2} \times m_{2} = \frac{16}{9}$
$m_{2} = \frac{32}{9}$
$∴$ Required equation of the diameter is-
$y = \frac{32 x}{9}$

Manipal UGET-201

Conic Section

119690 The polar equation of the circle with centre at $(2, \frac{π}{2})$ and radius 3 units is

1

r^{2} + 4 r \cos θ = 5

2

r^{2} + 4 r \sin θ = 5

3

r^{2} - 4 r \sin θ = 5

4

r^{2} - 4 r \cos θ = 5

Explanation:

C Given, point $(2, \frac{π}{2})$ and $r = 3$ units
We know that, general equation of the circle in polar form is -
$r^{2} - 2 {rr}_{0} \cos (θ - γ) + r_{0}^{2} = a^{2}$
Where $(r_{0}, γ)$ is the centre and $a$ is the radius
So, the equation of the circle with centre $(2, \frac{π}{2})$ and radius of 3 units is given by
$r^{2} - 2 r (2) \cos (θ - \frac{π}{2}) + (2)^{2} = (3)^{2}$
$r^{2} - 4 r \cos {- (\frac{π}{2} - θ)} + 4 = 9$
$r^{2} - 4 r \cos (\frac{π}{2} - θ) = 9 - 4$
$r^{2} - 4 r \sin θ = 9 - 4 {∵ \cos (\frac{π}{2} - θ) = \sin θ$
$r^{2} - 4 r \sin θ = 5$

Assam CEE-2022

Conic Section

119681 The equation of the circle which passes throug the points $(2, 3)$ and $(4, 5)$ and the centre lies 0 the straight line $y - 4 x + 3 = 0$ is

1

x^{2} + y^{2} + 4 x - 10 y + 25 = 0

2

x^{2} + y^{2} - 4 x - 10 y + 16 = 0

3

x^{2} + y^{2} - 4 x - 10 y + 25 = 0

4 None of the above

Explanation:

C Let, $(h, k)$ be the co-ordinate of centre and a be the radius of the circle. Then equation of circle
$(x - h)^{2} + (y - k)^{2} = a^{2}$
$∵ \dots . . (i)$
${CP}^{2} = {CQ}^{2}$
$\Rightarrow h - 2)^{2} + (k - 3)^{2} = (h - 4)^{2} + (k - 5)^{2}$
$\Rightarrow h + k = 7 \dots \dots . (ii)$
Also centre $(h, k)$ lies on the straight line
$∴ y - 4 x + 3 = 0$
$∴ - 4 h + k = - 3 .$
solving equation (2) and (3)
The co-ordinate of centre $= (2, 5)$
$∴ Radius (a) = \sqrt{(2 - 2)^{2} + (5 - 3)^{2}} = CP = CQ$
$= 2$
$∴$ from equation (i) equation of the circle is-
$(x - 2)^{2} + (y - 5)^{2} = 4$
or $x^{2} + y^{2} - 4 x - 10 y + 25 = 0$

Manipal UGET-201

Conic Section

119682 The point $(- 15, 21)$ lies
#[Qdiff: Hard, QCat: Numerical Based, examname: is, By comparing equation (i) with standard equation, $x^{2} + y^{2} = r^{2}$ , $r^{2} = 625$ , $r = 25$ , Distance of the point from the centre is, $d = \sqrt{(0 - (- 15)^{2} + (0 - 21)^{2}}$ , $= \sqrt{225 + 441} = 25.61 > 25$ Thus the point $(- 15, 21)$ lies out-side of circle, 65. Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its centre,

1 Outside the circle

x^{2} + y^{2} = 729

2 Outside the circle

x^{2} + y^{2} = 625

3 Inside the circle

x^{2} + y^{2} = 529

4 Inside the circle

x^{2} + y^{2} = 576

Explanation:

B Given point $(- 15, 21)$
Consider equation $\to x^{2} + y^{2} = 625$
Ans: c
Exp: (C) : Let c be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid-point of the chords of the circle $C$ that subtended an angle of $\frac{2 π}{3}$ at its centre.
In $△ CPB$
$\cos \frac{π}{3} = \frac{CP}{CB}$
$\frac{1}{2} = \frac{\sqrt{h^{2} + k^{2}}}{3}$ $h^{2} + k^{2} = \frac{9}{4}$
original image
Hence, The equation of the locus of the mid-points of the chords of the circle $C$ is $x^{2} + y^{2} = \frac{9}{4}$

is

Conic Section

119680 The diameter of $16 x^{2} - 9 y^{2} = 144$ which
conjugate to $x = 2 y$ is

1

y = \frac{32 x}{9}

2

x = \frac{16}{9} y

3

y = \frac{16}{9} x

4 None of these

Explanation:

A The diameters $y = m_{1} x$ and $y = m_{2} x$ are conjugate diameters of the hyperbola
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
Here $a^{2} = 9$ and $b^{2} = 16$
Also given $x = 2 y$
Then $m_{1} = \frac{1}{2}$
$∵ \frac{1}{2} \times m_{2} = \frac{16}{9}$
$m_{2} = \frac{32}{9}$
$∴$ Required equation of the diameter is-
$y = \frac{32 x}{9}$

Manipal UGET-201

Conic Section

119690 The polar equation of the circle with centre at $(2, \frac{π}{2})$ and radius 3 units is

1

r^{2} + 4 r \cos θ = 5

2

r^{2} + 4 r \sin θ = 5

3

r^{2} - 4 r \sin θ = 5

4

r^{2} - 4 r \cos θ = 5

Explanation:

C Given, point $(2, \frac{π}{2})$ and $r = 3$ units
We know that, general equation of the circle in polar form is -
$r^{2} - 2 {rr}_{0} \cos (θ - γ) + r_{0}^{2} = a^{2}$
Where $(r_{0}, γ)$ is the centre and $a$ is the radius
So, the equation of the circle with centre $(2, \frac{π}{2})$ and radius of 3 units is given by
$r^{2} - 2 r (2) \cos (θ - \frac{π}{2}) + (2)^{2} = (3)^{2}$
$r^{2} - 4 r \cos {- (\frac{π}{2} - θ)} + 4 = 9$
$r^{2} - 4 r \cos (\frac{π}{2} - θ) = 9 - 4$
$r^{2} - 4 r \sin θ = 9 - 4 {∵ \cos (\frac{π}{2} - θ) = \sin θ$
$r^{2} - 4 r \sin θ = 5$

Assam CEE-2022

Conic Section

119681 The equation of the circle which passes throug the points $(2, 3)$ and $(4, 5)$ and the centre lies 0 the straight line $y - 4 x + 3 = 0$ is

1

x^{2} + y^{2} + 4 x - 10 y + 25 = 0

2

x^{2} + y^{2} - 4 x - 10 y + 16 = 0

3

x^{2} + y^{2} - 4 x - 10 y + 25 = 0

4 None of the above

Explanation:

C Let, $(h, k)$ be the co-ordinate of centre and a be the radius of the circle. Then equation of circle
$(x - h)^{2} + (y - k)^{2} = a^{2}$
$∵ \dots . . (i)$
${CP}^{2} = {CQ}^{2}$
$\Rightarrow h - 2)^{2} + (k - 3)^{2} = (h - 4)^{2} + (k - 5)^{2}$
$\Rightarrow h + k = 7 \dots \dots . (ii)$
Also centre $(h, k)$ lies on the straight line
$∴ y - 4 x + 3 = 0$
$∴ - 4 h + k = - 3 .$
solving equation (2) and (3)
The co-ordinate of centre $= (2, 5)$
$∴ Radius (a) = \sqrt{(2 - 2)^{2} + (5 - 3)^{2}} = CP = CQ$
$= 2$
$∴$ from equation (i) equation of the circle is-
$(x - 2)^{2} + (y - 5)^{2} = 4$
or $x^{2} + y^{2} - 4 x - 10 y + 25 = 0$

Manipal UGET-201

Conic Section

119682 The point $(- 15, 21)$ lies
#[Qdiff: Hard, QCat: Numerical Based, examname: is, By comparing equation (i) with standard equation, $x^{2} + y^{2} = r^{2}$ , $r^{2} = 625$ , $r = 25$ , Distance of the point from the centre is, $d = \sqrt{(0 - (- 15)^{2} + (0 - 21)^{2}}$ , $= \sqrt{225 + 441} = 25.61 > 25$ Thus the point $(- 15, 21)$ lies out-side of circle, 65. Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the midpoints of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its centre,

1 Outside the circle

x^{2} + y^{2} = 729

2 Outside the circle

x^{2} + y^{2} = 625

3 Inside the circle

x^{2} + y^{2} = 529

4 Inside the circle

x^{2} + y^{2} = 576

Explanation:

B Given point $(- 15, 21)$
Consider equation $\to x^{2} + y^{2} = 625$
Ans: c
Exp: (C) : Let c be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid-point of the chords of the circle $C$ that subtended an angle of $\frac{2 π}{3}$ at its centre.
In $△ CPB$
$\cos \frac{π}{3} = \frac{CP}{CB}$
$\frac{1}{2} = \frac{\sqrt{h^{2} + k^{2}}}{3}$ $h^{2} + k^{2} = \frac{9}{4}$
original image
Hence, The equation of the locus of the mid-points of the chords of the circle $C$ is $x^{2} + y^{2} = \frac{9}{4}$

is

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