Conic Sections Test

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Conic Sections Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The centre of the circle given by $$\mathbf { r } \cdot ( \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } ) = 15 \text { and } | \mathbf { r } - ( \mathbf { j } + 2 \mathbf { k } ) | = 4 ,$$

A
$$( 0,1,2 )$$

B
$$( 1,3,4 )$$

C
$$( - 1,3,4 )$$

D
None of these

Solution

$$\begin{array}{l} The\, equation\, of\, line\, pas\sin g\, through\, centre \\ \vec { i } +2\vec { k } \, and\, normal\, to\, the\, given\, plane \\ \vec { r } =\vec { j } +2\vec { k } +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) \, \, \, \, \, \, -----\left( 1 \right) \\ This\, plane\, meets\, the\, plane\, at\, a\, po{ { int } }\, for\, which \\ \left[ { \left( { \vec { j } +2\vec { k } } \right) +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) } \right] \left( { \vec { i } +2\vec { j } +2\vec { k } } \right) =15 \\ \Rightarrow 6+9\lambda =15 \\ \Rightarrow 9\lambda =9 \\ \therefore \lambda =1 \\ substituting\, value\, of\, \lambda \, in\, eq{ u^{ n } }\left( 1 \right) \\ \vec { r } =\vec { j } +2\vec { k } +\vec { i } +2\vec { j } +2\vec { k } \\ \Rightarrow \vec { r } =\vec { i } +3\vec { j } +4\vec { k } \\ \therefore Centre\, is\, \left( { 1,\, 3,\, 4 } \right) \end{array}$$

Hence, option $$B$$ is the correct answer.
Question 2
1 / -0

The equation $${ x }^{ 2 }+{ y }^{ 2 }=9$$ meets x-axis at

A
$$(\pm 3,0)$$

B
$$(\pm 9,0)$$

C
$$(\pm 1,0)$$

D
(-5/14, 5/14)

Solution

$$x^2+y^2=9$$
The equation meets x-axis if the y-coordinate is $$0$$
So $$x^2+0=9\\x^2=9\\x=\sqrt9\\x=\pm 3$$
So the point is $$(\pm 3,0)$$
Question 3
1 / -0

Which is not represented by quadratic equation ?

A
Circle

B
Straight line

C
Parabola

D
Hyperbola

Solution

Equation of circle is given by:
$$x^2+y^2+2gx+2fy+c$$
Equation of straight line is given by:
$$y=mx+c$$
Equation of parabola is given by:
$$(y-k)^2=4p(x-h)^2$$
Equation of hyperbola is given by:
$$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$
Hence, we can see from the above equations that only equation of a straight line is not represented by quadratic equation.
Question 4
1 / -0

Find the area of $$x^2+y^2=49$$

A
154

B
49

C
88

D
None

Solution

Th eequation $$x^2+y^2=49$$ describes a circle with $$7$$ as radius
So the area is given as $$\pi r^2\\\dfrac{22}{7}\times 7^2=154$$
Question 5
1 / -0

If the the hyperbola $$\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$ passses though $$(4,3)$$

A
$${ b }^{ 2 }=3 $$

B
$${ b }^{ 2 }=9$$

C
$${ b }^{ 2 }=4$$

D
$${ b }^{ 2 }=100$$

Solution

Equation of hyer bola is $$\dfrac{x^2}{4}-\dfrac{y^2}{b^2}=1$$ passes through
$$(4,3)$$
So $$\dfrac{4^2}{4}-\dfrac{3^2}{b^2}=1\\4-\dfrac{9}{b^2}=1\\ \dfrac{9}{b^2} =3\implies b^2=3$$
Question 6
1 / -0

Radius of the circle $$2x^2+2y^2+8x+4y-3=0$$ is

A
$$\sqrt{17}$$

B
$$\sqrt{23}$$

C
$$\sqrt{\dfrac{13}{2}}$$

D
$$\sqrt{\dfrac{11}{2}}$$

Solution

Formula,

$$r=\sqrt{g^2+f^2-c}$$

From given, we have,

$$2g=8\Rightarrow g=4$$

$$2f=4\Rightarrow f=2$$

$$c=-3$$

$$r=\sqrt{4^2+2^2-(-3)}=\sqrt{23}$$
Question 7
1 / -0

The equation $$\dfrac {x^{2}}{2-r}+\dfrac {y^{2}}{r-5}+1=0$$ represents an ellipse, if

A
$$r > 2$$

B
$$2 < r < 5$$

C
$$r > 5$$

D
$$r \in (2,5)$$

Solution

Given $$\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$$ represents a ellipse
$$\implies \dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}=-1$$
$$\implies \dfrac{x^2}{r-2}+\dfrac{y^2}{5-r}=1$$
Since this equation is an ellipse so $$r-2>0,5-r>0\implies 2<r<5$$
Question 8
1 / -0

Find the value of a if $$y^2=4ax $$ pases through $$(8,8)$$

A
2

B
4

C
8

D
None

Solution

Given point $$(8,8)$$
Given equation $$y^2=4ax$$
$$\implies 8^2=4a(8)\\64=32a\\a=\dfrac{64}{32}\\a=2$$
Question 9
1 / -0

Find the Center of circle $$x^2+y^2-4x-8x+25=0$$

A
$$(2,4)$$

B
$$(-2,-4)$$

C
$$(4,2)$$

D
$$(-4,-2)$$

Solution

The general equation of center of circle $$x^2+y^2+2gx+2fy+c=0$$ is $$(-g,-f)$$

So the center of circle $$x^2+y^2-4x,-8x+25=0$$ is $$(2,4)$$
Question 10
1 / -0

The equation of the circle passing through $$(2,0)$$ and $$(0,4)$$ and having the minimum radius is ______________.

A
$${ x }^{ 2 }+{ y }^{ 2 }=4$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x+4y=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-x-2y=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }-2x-4y=0$$

Solution

Equation of circle with centre (a,b) and radius r,
$$\rightarrow (x-a)^2 + (y-b)^2 = r^2$$

For the circle to have minimum radius, given points must be diametric points,
So, Centre of circle = $$(\dfrac{2+0}{2},\dfrac{0+4}{2}) = (1,2)$$
Radius of circle = $$\sqrt{(1-0)^2 + (2-4)^2} = \sqrt{5}$$

So, The equation of circle is $$(x-1)^2+(y-2)^2 = 5$$
$$\rightarrow x^2+y^2-2x-4y=0$$

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Conic Sections Test - 13

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Conic Sections Test - 13

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Question 1

$$( 0,1,2 )$$

$$( 1,3,4 )$$

$$( - 1,3,4 )$$

None of these

Solution

Question 2

$$(\pm 3,0)$$

$$(\pm 9,0)$$

$$(\pm 1,0)$$

(-5/14, 5/14)

Solution

Question 3

Circle

Straight line

Parabola

Hyperbola

Solution

Question 4

154

49

88

None

Solution

Question 5

$${ b }^{ 2 }=3 $$

$${ b }^{ 2 }=9$$

$${ b }^{ 2 }=4$$

$${ b }^{ 2 }=100$$

Solution

Question 6

$$\sqrt{17}$$

$$\sqrt{23}$$

$$\sqrt{\dfrac{13}{2}}$$

$$\sqrt{\dfrac{11}{2}}$$

Solution

Question 7

$$r > 2$$

$$2 < r < 5$$

$$r > 5$$

$$r \in (2,5)$$

Solution

Question 8

2

4

8

None

Solution

Question 9

$$(2,4)$$

$$(-2,-4)$$

$$(4,2)$$

$$(-4,-2)$$

Solution

Question 10

$${ x }^{ 2 }+{ y }^{ 2 }=4$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+4y=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-x-2y=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x-4y=0$$

Solution

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