Conic Sections Test

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Conic Sections Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following equations of a circle has center at (1, -3) and radius of 5?

A
$$\displaystyle x^{2}+y^{2}=25$$

B
$$\displaystyle \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=25$$

C
$$\displaystyle \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

D
$$\displaystyle \left ( x+1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

Solution

the general equation of a circle with center at (a,b) and radius r is
$$(x-a)^2+(y-b)^2=r^2$$
so substituting the values we get the circle equation as
$$(x-1)^2+(y+3)^2=25$$ option B
Question 2
1 / -0

Determine the area enclosed by the curve $$\displaystyle x^{2}-10x+4y+y^{2}=196$$

A
$$15\pi$$

B
$$225\pi$$

C
$$20\pi$$

D
$$17\pi$$

Solution

Equation of the circle is
$$x^2-10x+4y+y^2=196$$
$$=>x^2-2\times x \times 5 +5^2+y^2+2\times y\times 2+2^2=196+2^2+5^2$$
$$=>(x-5)^2+(y+2)^2=196+4+25$$
$$=>(x-5)^2+(y+2)^2=225$$
$$=>(x-5)^2+(y+2)^2=15^2$$
Thus radius of the circle is 15
Thus, area of the circle $$=\pi \times 15^2$$
$$=225\pi$$
Question 3
1 / -0

The standard equation of circle at origin is

A
$$x^2+y^2=r^2$$

B
$$x^2-y^2=r$$

C
$$x^2+y^2=1$$

D
$$x^2+y^2=0$$

Solution

The standard equation of circle at origin is $$x^2+y^2=r^2$$
Example: A circle with its center at the origin $$(0, 0)$$, and radius $$r$$ has the equation $$x^2 + y^2 = r^2$$. We'll know if a given point is on the circle if the coordinates of that point satisfy the equation.
Question 4
1 / -0

The diameter of a circle described by $$\displaystyle 9x^{2}+9y^{2}=16$$ is

A
$$\dfrac {16}9$$

B
$$\dfrac 43$$

C
$$4$$

D
$$\dfrac 83$$

Solution

Equation of circle is
$$9x^2+9y^2=16$$
$$=>x^2+y^2=\dfrac{16}{9}$$
$$=>x^2+y^2=(\dfrac{4}{3})^2$$
$$\therefore$$ Radius of the circle =$$\dfrac{4}{3}$$
$$\therefore$$ Diameter of the circle=$$\dfrac{4\times 2}{3}$$
$$=\dfrac{8}{3}$$
Question 5
1 / -0

A circle has a diameter whose ends are at (-3, 2) and (12, -6) Its Equation is

A
$$\displaystyle 4x^{2}+4y^{2}-36x+16y+192=0$$

B
$$\displaystyle 4x^{2}+4y^{2}-36x+16y-192=0$$

C
$$\displaystyle 4x^{2}+4y^{2}-36x-16y-192=0$$

D
$$\displaystyle 4x^{2}+4y^{2}-36x-16y+192=0$$

Solution

Coordinate of diameter is $$(-3,2)$$ and $$(12,-6)$$
Centre of the circle is the midpoint of the diameter.
$$\therefore (x_1,y_1)=(\dfrac{-3+12}{2}, \dfrac{2-6}{2})$$
$$=(\dfrac{9}{2}, \dfrac{-4}{2})$$
$$=(\dfrac{9}{2}, -2)$$
Half the distance between $$(-3,2)$$ and $$(12,-6)$$ is the radius.
$$\therefore 2r=\sqrt {(-3-12)^2+(2+(-6))^2}$$
$$=>2r=\sqrt {15^2+8^2}$$
$$=>2r=\sqrt {225+64}$$
$$=>2r=17$$
$$=>r=\dfrac{17}{2}$$
$$\therefore Equation of circle is
$$x-(\dfrac{9}{2})^2+(y-(-2))^2=(\dfrac{17}{2})^2$$
$$=>\dfrac{(2x-9)^2}{4}+(y+2)^2=\dfrac{289}{4}$$
$$=>4x^2-36x+81+4(y^2+4y+4)-289=0$$
$$=>4x^2-36x+81+4y^2+16y+16-289=0$$
$$=>4x^2+4y^2-36x+16y+192=0$$
Question 6
1 / -0

Find the equation of a circle with center $$(0, 0)$$ and radius $$5$$.

A
$$x^2+y^2=5$$

B
$$x^2-y^2=25$$

C
$$x^2+y^2=25$$

D
$$(x-1)^2+(y+1)^2=25$$

Solution

Compare the equation with the standard form with center at $$(h, k)$$ and radius $$r$$ is: $$(x-h)^2+(y-k)^2=r^2$$
Substitute the value of $$(h, k) = (0, 0)$$ and $$r = 5$$
Then, the equation becomes $$x^2+y^2 = 25$$
Question 7
1 / -0

What is the radius of the circle with the following equation?
$$\displaystyle x^{2}-6x+y^{2}-4y-12=0$$

A
$$3.46$$

B
$$5$$

C
$$7$$

D
$$6$$

Solution

Equation of circle is
$$x^2-6x+y^2-4y-12=0$$
$$=>x^2-2.3.x+y^2-2.2y-12=0$$
$$=>x^2-2.3.x+3^2+y^2-2.2.y+2^2=12+3^2+2^2$$
$$=>(x-3)^2+(y-2)^2=25$$
$$=>(x-3)^2+(y-2)^2=5^2$$
$$\therefore$$ Radius of the circle=$$5$$
Question 8
1 / -0

The locus of a planet orbiting around the sun is:

A
A circle

B
A straight line

C
A semicircle

D
An ellipse

Solution

It is a fact & proof of it can be seen from higher education physics books
Question 9
1 / -0

Number of intersecting points of the conic $$4x^{2} + 9y^{2} = 1$$ and $$4x^{2} + y^{2} = 4$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$0$$ (zero)

Solution

Number of intersecting points of the conic
Lets assume:
$$C_1: 4x^{2} + 9y^{2} = 1$$
$$C_2: 4x^{2} + y^{2} = 4$$

They can be written as:
$$C_{1} : \dfrac {x^{2}}{\dfrac {1}{4}} + \dfrac {y^{2}}{\dfrac {1}{9}} = 1$$

$$C_{2} : \dfrac {x^{2}}{1} + \dfrac {y^{2}}{4} = 1$$

Plotting these on coordinate axes, we get the following figure.
We can see that there is no intersection.
Hence, zero point of intersections.
Question 10
1 / -0

What is the nature of the given graph?

A
The graph in symmetric about x-axis

B
The graph in symmetric about y-axis

C
Sum of x-intercept and y-intercept is greater than zero

D
The polynomial has $$3$$ terms

Solution

R.E.F image
As graph is touching x axis
& graph is increasing on both
side about is axis $$ \therefore $$ has 2
equal roots at $$ \boxed{x = 0};$$ hence
$$ 2^{\circ}$$ Polynomial.
$$ \Rightarrow y = ax^{2}+bx+c $$
$$ \Rightarrow \boxed{y = x^{2}} $$ using $$ \Rightarrow $$ $$\begin{bmatrix} At\,x = 0 ; y =0 \\ x=1;y=1 \\ x=-1;y=1 & \end{bmatrix}$$

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Conic Sections Test - 10

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Conic Sections Test - 10

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Question 1

$$\displaystyle x^{2}+y^{2}=25$$

$$\displaystyle \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=25$$

$$\displaystyle \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

$$\displaystyle \left ( x+1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

Solution

Question 2

$$15\pi$$

$$225\pi$$

$$20\pi$$

$$17\pi$$

Solution

Question 3

$$x^2+y^2=r^2$$

$$x^2-y^2=r$$

$$x^2+y^2=1$$

$$x^2+y^2=0$$

Solution

Question 4

$$\dfrac {16}9$$

$$\dfrac 43$$

$$4$$

$$\dfrac 83$$

Solution

Question 5

$$\displaystyle 4x^{2}+4y^{2}-36x+16y+192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x+16y-192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x-16y-192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x-16y+192=0$$

Solution

Question 6

$$x^2+y^2=5$$

$$x^2-y^2=25$$

$$x^2+y^2=25$$

$$(x-1)^2+(y+1)^2=25$$

Solution

Question 7

$$3.46$$

$$5$$

$$7$$

$$6$$

Solution

Question 8

A circle

A straight line

A semicircle

An ellipse

Solution

Question 9

$$1$$

$$2$$

$$3$$

$$0$$ (zero)

Solution

Question 10

The graph in symmetric about x-axis

The graph in symmetric about y-axis

Sum of x-intercept and y-intercept is greater than zero

The polynomial has $$3$$ terms

Solution

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