Conic Sections Test

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Conic Sections Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

What is the approximate radius of the circle whose equation is $$(x-\sqrt{3})^2+(y+2)^2=11$$?

A
1.71

B
2.33

C
3.32

D
3.85

E
4.27
Solution
- The radius of given circle is $$\sqrt{11} = 3.32$$
Question 2
1 / -0

Which of the following is an equation of a circle in the $$xy$$-plane with center $$\left(0, 4\right)$$ and a radius with endpoint $$\left(\dfrac{4}{3}, 5\right)$$?

A
$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

B
$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

C
$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 5 }{ 3 } $$

D
$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 3 }{ 5 } $$

Solution

Given, center of circle is $$(0,4)$$ and point on circle $$\left (\dfrac 43,5\right)$$
$$\therefore$$ radius of circle $$=\sqrt {\left (\dfrac 43-0\right)^2+(5-4)^2}$$
$$= \sqrt {\dfrac {16}{9}+1}=\sqrt {\dfrac {25}9}$$
Since, Equation of the circle having centre $$(x_1,y_1)$$ and radius '$$r$$' is
$$(x-x_1)^2+(y-y_1)^2=r^2$$
$$\therefore$$ Equation of given circle $$= (x-0)^2+(y-4)^2=\left (\sqrt {\dfrac {25}9}\right)^2$$
$$= x^2+(y-4)^2=\dfrac {25}9$$
Hence, option A is correct.
Question 3
1 / -0

Identify the polynomial represented by the graph?

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

B
$$y = -x^{2}$$

C
$$y = x^{2}$$

D
$$y = x^{2} - 2$$

Solution
Question 4
1 / -0

The least value of $$2x^{2} + y^{2} + 2xy + 2x - 3y + 8$$ for real numbers $$x$$ and $$y$$ is

A
$$2$$

B
$$8$$

C
$$3$$

D
$$1$$

E
$$-\dfrac12$$

Solution

$$E = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$$

$$\Rightarrow E = \dfrac {1}{2} (4x^{2} + 2y^{2} + 4xy + 4x - 6y + 16)$$

$$\Rightarrow E= \dfrac {1}{2}\left \{(y - 4)^{2} + (2x + y + 1)^{2} - 1\right \} \geq - \dfrac {1}{2}$$

This happen when the $$\left \{(y - 4)^{2} + (2x + y + 1)^{2} - 1\right \} = -1$$
Which i possible when $$(x,y) \equiv \left(-\dfrac52,4\right)$$

So, least value is $$-\dfrac {1}{2}$$
Question 5
1 / -0

Locus of the point $$(\sqrt{3h} , \sqrt{3k + 2} )$$ if it lies on the line $$x- y- 1 = 0$$ is a

A
Straight line

B
Circle

C
Parabola

D
None of these

Solution

Locus of point $$\left( \sqrt { 3h } ,\sqrt { 3k+2 } \right) $$ if it lies on line $$x-y-1=0$$ is :
$$\sqrt { 3 } h-\sqrt { 3k+2 } -1=0\\ { (\sqrt { 3 } h-1) }^{ 2 }={ (\sqrt { 3k+2 } ) }^{ 2 }\\ { 3k }^{ 2 }+1-2\sqrt { 3 } h=3k+2\\ \left( h-\cfrac { 1 }{ \sqrt { 3 } } \right) ^{ 2 }=3\left( k+\cfrac { 2 }{ 9 } \right) \\ x^{ 2 }=3y$$
Which is parabola.
Question 6
1 / -0

The length of the latus rectum of the parabola whose vertex is $$(2, -3)$$ and the directrix $$x = 4$$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

Distance of vertex $$(2,-3)$$ from directrix $$x=4$$ is
$$\quad \quad =\left|\dfrac{2-4}{\sqrt{1^2+0^2}} \right|=2$$
So length of latue rectum of above parabola is
$$=4\times $$ distance of vertex to directrix $$=4\times 2=8$$
Question 7
1 / -0

The graph of the equation $$x^2+\dfrac{y^2}{4}=1$$ is

A
an ellipse

B
a circle

C
a hyperbola

D
a parabola

E
two straight lines

Solution

Given, $${x}^{2}+\dfrac{{y}^{2}}{4}=1$$
It is in the form of $$ \dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$$
Therefore, it represents an ellipse.
Question 8
1 / -0

Which ordered number pair represents the center of the circle $$x^2 + y^2 - 6x + 4y - 12 = 0$$?

A
(9,4)

B
(3,2)

C
(3,-2)

D
(6,4)

Solution

The equation of circle is $$x^{ 2 }+y^{ 2 }-6x+4y-12=0$$
$$\Rightarrow x^2-6x+9+y^2+4y+4=12+13$$
$$\Rightarrow (x-3)^{ 2 }+(y+2)^{ 2 }=25$$
Comparing above equation with $$(x-h)^2+(y-k)^2=r^2$$
Therefore, we get the center of circle as $$(3,-2)$$.
Question 9
1 / -0

Equation of circle with center (-a, -b) and radius $$\sqrt{a^2-b^2}$$ is.

A
$$x^2+y^2-2ax-2by - 2b^2=0$$

B
$$x^2+y^2-2ax-2by + 2a^2=0$$

C
$$x^2+y^2+2ax + 2by + 2b^2=0$$

D
$$x^2+y^2-2ax-2by + 2b^2=0$$

Solution

Fact: equation of circle with centre $$(h,k)$$ and radius $$r$$ is given by

$$(x-h)^2+(y-k)^2=r^2$$

Hence equation of circle with centre $$(-a,-b)$$ having radius $$\sqrt{a^2-b^2}$$ is given by,

$$(x+a)^2+(y+b)^2=a^2-b^2$$

$$\Rightarrow x^2+y^2+2ax+2by+a^2+b^2=a^2-b^2$$

$$\Rightarrow x^2+y^2+2ax+2by+2b^2=0$$
Question 10
1 / -0

The asymptotes of a hyperbola $$4x^2 - 9y^2=36$$ are

A
$$2x \pm 3y = 1$$

B
$$2x \pm 3y = 0$$

C
$$3x \pm 2y = 1$$

D
None

Solution

The equation of hyperbola is $$\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$
So the equation of asymptotes is $$\displaystyle \frac{x^{2}}{9}-\frac{y^{2}}{4}=0$$
$$\Rightarrow 4x^{2}-9y^{2}=0$$
$$\Rightarrow 2x \pm 3y=0$$
Therefore option $$B$$ is correct

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Conic Sections Test - 24

Result

Conic Sections Test - 24

Score

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Rank

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SHARING IS CARING

Question 1

1.71

2.33

3.32

3.85

4.27

Solution

Question 2

$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 5 }{ 3 } $$

$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 3 }{ 5 } $$

Solution

Question 3

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

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

$$y = x^{2}$$

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

Solution

Question 4

$$2$$

$$8$$

$$3$$

$$1$$

$$-\dfrac12$$

Solution

Question 5

Straight line

Circle

Parabola

None of these

Solution

Question 6

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 7

an ellipse

a circle

a hyperbola

a parabola

two straight lines

Solution

Question 8

(9,4)

(3,2)

(3,-2)

(6,4)

Solution

Question 9

$$x^2+y^2-2ax-2by - 2b^2=0$$

$$x^2+y^2-2ax-2by + 2a^2=0$$

$$x^2+y^2+2ax + 2by + 2b^2=0$$

$$x^2+y^2-2ax-2by + 2b^2=0$$

Solution

Question 10

$$2x \pm 3y = 1$$

$$2x \pm 3y = 0$$

$$3x \pm 2y = 1$$

None

Solution

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