Conic Sections Test

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Conic Sections Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The point $$(3, 4)$$ is the focus and $$2x - 3y + 5 = 0$$ is the directrix of a parabola. Lenghth of latus rectum is

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

B
$$\dfrac {4}{\sqrt {13}}$$

C
$$\dfrac {1}{\sqrt {13}}$$

D
$$\dfrac {3}{\sqrt {13}}$$

Solution

We know that, Latus rectum $$= 2\times $$ (distance from focus to directrix)
$$= 2\cdot \dfrac {|6 - 12 + 5|}{\sqrt {4 + 9}} = \dfrac {2}{\sqrt {13}}$$
Question 2
1 / -0

If the centre $$O$$ of circle is the intersection of $$x-$$axis and line $$y=\dfrac { 4 }{ 3 } x+4$$, and the point $$(3,8)$$ lies on circle, then the equation of circle will be

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

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

C
$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=100$$

D
$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=100$$

Solution

The intersection of $$x$$-axis and the line $$y = \cfrac{4x}{3} + 4$$ is $$(-3,0)$$, which is the centre of the circle.
Since the point $$(3,8)$$ lies on the circle, its radius becomes the distance between these points, which is $$\sqrt{(3 + 3)^2 + (8 - 0)^2} = 10$$
The equation thus becomes $$(x + 3)^2 + (y - 0)^2 = (10)^2$$
i.e. $$(x + 3)^2 + y^2 = 100$$
Question 3
1 / -0

If the straight line $$y=mx+c$$ is parallel to the axis of the parabola $$y^2=lx$$ and intersects the parabola at $$\left(\dfrac{c^2}{8}, c\right)$$ then the length of the latus rectum is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$8$$

Solution

Length of Latus rectum of the parabola $$y^2=4ax$$ is $$4|a|$$
Here, parabola given is $$y^2 = lx$$
So, length of latus rectum will be $$l$$
It is also given that a straight line $$y = mx+c$$ intersect the given parabola at $$\left(\dfrac{c^2}8 ,c\right)$$
Therefore, this point $$\left(\dfrac{c^2}8 ,c\right)$$ should satisfy the equation of parabola
$$y^2 = lx$$
$$\because y = c, x = \dfrac{c^2}8$$
$$\Rightarrow c^2 = l\times\dfrac{c^2}8$$
$$\Rightarrow l = 8$$

$$\therefore$$ The length of latus rectum will be $$l = 8$$
Question 4
1 / -0

The area of the circle represented by the equation $${(x+3)}^{2}+{(y+1)}^{2}=25$$ is

A
$$4\pi$$

B
$$5\pi$$

C
$$16\pi$$

D
$$25\pi$$

Solution

The general equation of circle with centre $$(h,k)$$ and radius $$r$$ is given by $${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ \left( r \right) }^{ 2 }$$.
It is given that the equation of the circle is $$(x+3)^2+(y+1)^2=25$$
Therefore, $$h=-3$$, $$k=-1$$ and $$r=5$$.
The area of the circle with radius $$r$$ is $$\pi { r }^{ 2 }$$, thus the area of the circle with radius $$r=5$$ is:
$$\pi { r }^{ 2 }=\pi { (5) }^{ 2 }=25\pi$$
Hence, the area of the circle is $$25\pi$$.
Question 5
1 / -0

The radius of the circle passing through the point $$(6, 2)$$ and two of whose diameters are $$\displaystyle x+y=6$$ and $$\displaystyle x+2y=4$$ is:

A
$$4$$

B
$$6$$

C
$$20$$

D
$$\displaystyle \sqrt { 20 } $$

Solution

Point of intersection of the given diameters is $$(8,-2)$$ which is the centre of the circle
Also the circle pass through the point $$(6,2)$$, so the radius is $$=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{20}$$
Question 6
1 / -0

Write the equation of the circle with center at $$(0,0)$$ and a radius of $$6$$

A
$$(x-6)^2+y^2=36$$

B
$$x^2+(y-6)^2=0$$

C
$$x^2+y^2=36$$

D
$$x^2+y^2=-36$$

Solution

An equation of the circle with center $$(h,k)$$ and radius $$r$$ is
$$(x-h)^{ 2 }+(y-k)^{ 2 }=r^{ 2 }$$
So, if the center is $$(0,0)$$ and the radius is $$6$$, an equation of the circle is: $$(x-0)^{ 2 }+(y-0)^{ 2 }=6^{ 2 }$$
On simplifying, we get:
$$x^{ 2 }+y^{ 2 }=36$$
Hence, the equation of the circle is $$x^{ 2 }+y^{ 2 }=36$$.
Question 7
1 / -0

The graph of the equation $$x^2+2y^2

= 8$$ is

A
a circle

B
an ellipse

C
a hyperbola

D
a parabola

Solution
Question 8
1 / -0

The graph of the equation $$4y^2 + x^2= 25$$ is

A
a circle

B
an ellipse

C
a hyperbola

D
a parabola

E
a straight line

Solution

Given, $$4{y}^{2}+{x}^{2}=25$$
$$\Rightarrow \dfrac { { y }^{ 2 } }{ 25/4 } +\dfrac { { x }^{ 2 } }{ 25 } =1$$
It is in the form of ellipse $$\left (\dfrac { { y }^{ 2 } }{ {a}^{2} } +\dfrac { { x }^{ 2 } }{ {b}^{2} } =1\right)$$
So, the correct answer is option $$B$$.
Question 9
1 / -0

Which of the following is an equation of the circle with its center at $$(0,0)$$ that passes through $$(3,4)$$ in the standard $$(x,y)$$ coordinate plane?

A
$$x+y=1$$

B
$$x-y=25$$

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

D
$${x}^{2}+{y}^{2}=5$$

E
$${x}^{2}+{y}^{2}=25$$

Solution

Let the radius of the circle be '$$r$$'
Then the equation of the circle is
$$(x-0)^{2}+(y-0)^{2}=r^{2}$$
Hence $$x^{2}+y^{2}=r^{2}$$
Now, it passes through $$(3,4)$$.
Hence $$3^{2}+4^{2}=r^{2}$$ $$\Rightarrow r^{2}=25$$.
Therefore, the equation of the circle is $$x^{2}+y^{2}=25$$.
Question 10
1 / -0

A circle with center $$(3, 8)$$ contains the point $$(2, -1)$$. Another point on the circle is:

A
$$(1, -10)$$

B
$$(4, 17)$$

C
$$(5, -9)$$

D
$$(7, 15)$$

E
$$(9, 6)$$

Solution

$$(x-3)^2+(y-8)^2=r^2$$
Given $$(2,-1)$$ lies on the circle
$$(2-3)^2+(-1-8)^2=r^2$$
$$1+81=r^2$$
$$r^2=82$$
Circle equation is :$$(x-3)^2+(y-8)^2=82$$
By trial and error, substitute the point
in the above equation
$$(4-3)^2+(17-8)^2=82$$
hence , $$(4,17)$$ satisfy the circle euation.

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 23

Result

Conic Sections Test - 23

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SHARING IS CARING

Question 1

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

$$\dfrac {4}{\sqrt {13}}$$

$$\dfrac {1}{\sqrt {13}}$$

$$\dfrac {3}{\sqrt {13}}$$

Solution

Question 2

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

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

$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=100$$

$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=100$$

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$8$$

Solution

Question 4

$$4\pi$$

$$5\pi$$

$$16\pi$$

$$25\pi$$

Solution

Question 5

$$4$$

$$6$$

$$20$$

$$\displaystyle \sqrt { 20 } $$

Solution

Question 6

$$(x-6)^2+y^2=36$$

$$x^2+(y-6)^2=0$$

$$x^2+y^2=36$$

$$x^2+y^2=-36$$

Solution

Question 7

a circle

an ellipse

a hyperbola

a parabola

Solution

Question 8

a circle

an ellipse

a hyperbola

a parabola

a straight line

Solution

Question 9

$$x+y=1$$

$$x-y=25$$

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

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

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

Solution

Question 10

$$(1, -10)$$

$$(4, 17)$$

$$(5, -9)$$

$$(7, 15)$$

$$(9, 6)$$

Solution

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