Conic Sections Test

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Conic Sections Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

Vertex of the parabola $$9x^2 - 6x + 36y + 9 = 0$$ is

A
$$\bigg(\dfrac{1}{3},\dfrac{-2}{9}\bigg)$$

B
$$\bigg(\dfrac{-1}{3},\dfrac{1}{2}\bigg)$$

C
$$\bigg(\dfrac{-1}{3},\dfrac{-1}{2}\bigg)$$

D
$$\bigg(\dfrac{1}{3},\dfrac{1}{2}\bigg)$$

Solution

Given,

$$9x^2-6x+36y+9=0$$

$$(3x)^2-2(3x)=-36y-9$$

$$(3x)^2-2(3x)+1=-36y-9+1$$

$$(3x-1)^2=-36y-8$$

$$(3x-1)^2=-36\left ( y+\dfrac{2}{9} \right )$$

$$(x-a)^2=4c(y-b)$$

$$\left ( x-\dfrac{1}{3} \right )^2=-4\left ( y+\dfrac{2}{9} \right )$$

$$x-\dfrac{1}{3}=0$$

$$\Rightarrow x=\dfrac{1}{3}$$

$$y+\dfrac{2}{9} =0$$

$$\Rightarrow y=-\dfrac{2}{9}$$

$$\therefore \left ( \dfrac{1}{3}, -\dfrac{2}{9}\right )$$
Question 2
1 / -0

If two vertices of an equilateral triangle are $$A (-a, 0)$$ and $$B(a, 0), a > 0$$ and the third vertex C lies above x-axis then the equation of the circumcircle of $$\Delta ABC$$ is

A
$$3x^2+3y^2-2\sqrt{3} ay =3a^2$$

B
$$3x^2+3y^2-2 ay =3a^2$$

C
$$x^2+y^2-2ay =a^2$$

D
$$x^2+y^2-\sqrt{3} ay =3a^2$$

Solution
Question 3
1 / -0

The equation(s) of the circle(s) which pass through the ends of the common chords of two circles $$2x^{2}+2y^{2}+8x+4y-7=0$$ and $$x^{2}+y^{2}-8x-4y-5=0$$ and touch the line $$x=7$$ is (are) :

A
$$x^{2}+y^{2}-6x+2y-\dfrac{19}{4}=0$$

B
$$x^{2}+y^{2}+120x+60y+11=0$$

C
$$x^{2}+y^{2}-6x+2y+\dfrac{19}{4}=0$$

D
$$x^{2}+y^{2}+120x+60y-11=0$$
Question 4
1 / -0

The equation of the circle which touches the axes of $$y$$ at the origin and passing through $$(3,4)$$ is

A
$$4(x^{2}+y^{2})-25x=0$$

B
$$3(x^{2}+y^{2})-25x=0$$

C
$$2(x^{2}+y^{2})-3x=0$$

D
$$4(x^{2}+y^{2})-25x=0$$

Solution
Question 5
1 / -0

The locus of center of a variable circle touching the circle of radius $${ r }_{ 1 }and{ r }_{ 2 }$$ extemally which also touch each other externally , is a conic of the eccentricity $$e$$.If $$\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =3+2\sqrt { 2 } $$ then $${ e }^{ 2 }$$ is

A
2

B
3

C
4

D
5

Solution
Question 6
1 / -0

The value of k, such that the equation
$$2x^{2}+2y^{2}-6x+8y+k=0$$ represent a point circle , is

A
0

B
25

C
$$\frac {25}{2}$$

D
$$-\frac {25}{2}$$

Solution

The given equation of circle is
$$\Rightarrow$$ $$2x^2+2y^2-6x+8y+k=0$$
$$\Rightarrow$$ $$x^2+y^2-3x+4y+\dfrac{k}{2}=0$$ [ Dividing both sides by $$2$$ ]
Comparing it with general equation of circle $$x^2+y^2+2gx+2fy+c=0,$$ we get
$$\Rightarrow$$ $$2g=-3,\,2f=4$$ and $$c=\dfrac{k}{2}$$
$$\therefore$$ $$g=\dfrac{-3}{2}$$ and $$f=2$$
Center $$=(-g,-f)=\left(\dfrac{3}{2},\,-2\right)$$
We know,
Radius $$(r)=\sqrt{g^2+f^2-c}$$
We know that for point circle radius is $$0.$$ $$r=0$$
$$\Rightarrow$$ $$0=\sqrt{\left(\dfrac{3}{2}\right)^2+(-2)^2-\dfrac{k}{2}}$$
$$\Rightarrow$$ $$0=\sqrt{\dfrac{9}{4}+4-\dfrac{k}{2}}$$
$$\Rightarrow$$ $$0=\dfrac{9}{4}+4-\dfrac{k}{2}$$
$$\Rightarrow$$ $$\dfrac{k}{2}=\dfrac{9+16}{4}$$
$$\Rightarrow$$ $$\dfrac{k}{2}=\dfrac{25}{4}$$
$$\therefore$$ $$k=\dfrac{25}{2}$$
Question 7
1 / -0

If the radius of the circle $$x^{2}+y^{2}-18x-12y+k=0$$ be $$11$$ then k=

A
34

B
4

C
-4

D
49

Solution

The equation of circle is $$x^2+y^2-18x-12y+k=0$$.
Comparing it with $$x^2+y^2+2gx+2fy+c=0$$ we get,
$$2g=-18,\,2f=-12$$ and $$c=k$$
$$\therefore$$ $$g=-9,\,f=-6$$ and $$c=k$$
Center $$=(-g,-f)=(9,6)$$
Radius of a circle $$(r)=11$$ [ Given ]
We know,
$$\Rightarrow$$ $$r=\sqrt{g^2+f^2-c}$$
$$\Rightarrow$$ $$11=\sqrt{(-9)^2+(-6)^2-k}$$
$$\Rightarrow$$ $$11=\sqrt{81+36-k}$$
$$\Rightarrow$$ $$11=\sqrt{117-k}$$
$$\Rightarrow$$ $$121=117-k$$
$$\Rightarrow$$ $$4=-k$$
$$\therefore$$ $$k=-4$$
Question 8
1 / -0

The equation of a circle which is passing through the vertices of an equilateral triangle whose median is of length 3 a is

A
$$

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

$$

B
$$

x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }

$$

C
$$

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

$$

D
$$

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

$$

Solution

Median of the equilateral triangle is $$3a$$
$$\Rightarrow=3a$$
In $$\triangle ADB$$,we have
$$\Rightarrow(AB)^2=(AD)^2+(BD)^2$$
$$\Rightarrow(AB)^2=(3a)^2+\left(\dfrac{AB}{2}\right)^2$$
$$\Rightarrow(AB)^2=9a^2+\dfrac{(AB)^2}{4}$$
$$\Rightarrow\dfrac{3}{4}(AB)^2=9a^2$$
$$\Rightarrow(AB)^2=12a^2$$
Now, In $$\triangle OBD,$$
$$\Rightarrow(OB)^2=(OD)^2+(BD)^2$$
$$\Rightarrow r^2=(3a-r)^2+\left(\dfrac{AB}{2}\right)^2$$
$$\Rightarrow r^2=9a^2+r^2-6ar+3a^2$$ $$[\because(AB)^2=10a^2]$$
$$\Rightarrow6ar=12a^2$$
$$\Rightarrow r=2a$$
So, the equation of circle is
$$\Rightarrow(x-0)^2+(y-0)^2=(2a^2)$$
$$\Rightarrow x^2+y^2=4a^2$$
Question 9
1 / -0

Equation $$\sqrt{(x-2)^2 + y^2} + \sqrt{(x+2)^2 + y^2}$$ = 4 represents _____________.

A
Parabola

B
Ellipse

C
Circle

D
straight lines

Solution
Question 10
1 / -0

If $$\left| z-{ z }_{ 0 } \right| =\dfrac { \left| a\overline { z } +a\overline { z } +b \right| }{ 2\left| a \right| } $$ represent a parabola, Then the length of latus rectum of parabola

A
$$\left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| $$

B
$$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ \left| a \right| } $$

C
$$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ 2\left| a \right| } $$

D
None of these

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Conic Sections Test - 47

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Conic Sections Test - 47

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SHARING IS CARING

Question 1

$$\bigg(\dfrac{1}{3},\dfrac{-2}{9}\bigg)$$

$$\bigg(\dfrac{-1}{3},\dfrac{1}{2}\bigg)$$

$$\bigg(\dfrac{-1}{3},\dfrac{-1}{2}\bigg)$$

$$\bigg(\dfrac{1}{3},\dfrac{1}{2}\bigg)$$

Solution

Question 2

$$3x^2+3y^2-2\sqrt{3} ay =3a^2$$

$$3x^2+3y^2-2 ay =3a^2$$

$$x^2+y^2-2ay =a^2$$

$$x^2+y^2-\sqrt{3} ay =3a^2$$

Solution

Question 3

$$x^{2}+y^{2}-6x+2y-\dfrac{19}{4}=0$$

$$x^{2}+y^{2}+120x+60y+11=0$$

$$x^{2}+y^{2}-6x+2y+\dfrac{19}{4}=0$$

$$x^{2}+y^{2}+120x+60y-11=0$$

Question 4

$$4(x^{2}+y^{2})-25x=0$$

$$3(x^{2}+y^{2})-25x=0$$

$$2(x^{2}+y^{2})-3x=0$$

$$4(x^{2}+y^{2})-25x=0$$

Solution

Question 5

2

3

4

5

Solution

Question 6

0

25

$$\frac {25}{2}$$

$$-\frac {25}{2}$$

Solution

Question 7

34

4

-4

49

Solution

Question 8

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

$$x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }$$

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

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

Solution

Question 9

Parabola

Ellipse

Circle

straight lines

Solution

Question 10

$$\left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| $$

$$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ \left| a \right| } $$

$$\dfrac { \left| a\overline { { z }_{ 0 } } \overline { { a }z_{ 0 } } +b \right| }{ 2\left| a \right| } $$

None of these

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$$

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

$$

$$

x ^ { 2 } + y ^ { 2 } = 16 a ^ { 2 }

$$

$$

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

$$

$$

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

$$