Conic Sections Test

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Conic Sections Test - 25

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Weekly Quiz Competition

Question 1
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The equation of hyperbola whose coordinates of the foci are $$(\pm8,0)$$ and the lenght of latus rectum is $$24$$ units, is

A
$$3{ x }^{ 2 }-{ y }^{ 2 }=48$$

B
$$4{ x }^{ 2 }-{ y }^{ 2 }=48$$

C
$${ x }^{ 2 }-3{ y }^{ 2 }=48$$

D
$${ x }^{ 2 }-4{ y }^{ 2 }=48$$

Solution

The Foci of hyperbola are $$(\pm8,0)$$, hence Foci lie on $$x$$ - axis.

We know that foci of hyperbola lie at $$(\pm ae,0)$$, So $$ae = 8$$ ...$$(1)$$

squaring both sides of equation $$(1)$$, we get,

$$\Rightarrow a^2e^2 = 64$$

Eccentricity of hyperbola $$e^2 =1 + \dfrac {b^2}{a^2}$$

$$\Rightarrow a^2(1+\dfrac{b^2}{a^2}) = 64$$

$$\Rightarrow a^2 + b^2 = 64$$ ...$$(2)$$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola $$ = \dfrac{2b^2}{a} = 24$$

$$\Rightarrow b^2 = 12a$$ ...$$(3)$$

putting value of $$b^2$$ in eq. $$(2)$$, we get,

$$\Rightarrow a^2 +12a -64 = 0$$

Hence $$a = 4, -16$$

As $$a$$ is always taken as positive value so $$a =4$$

from eq. $$(3)$$, $$b = \sqrt{48}$$

Hence equation of hyperbola is $$\dfrac{x^2}{16} -\dfrac {y^2}{48} = 1$$

Or $$3x^2 -y^2 = 48$$, So correct option is $$A$$.
Question 2
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The lines $$2x - 3y - 5 = 0$$ and $$3x -4y = 7$$ are diameters of a circle of area 154 sq units, then the equation of the circle is.( Use $$\pi = \dfrac{22}{7}$$)

A
$$x^2+y^2+2x-2y-62=0$$

B
$$x^2+y^2+2x-2y-47=0$$

C
$$x^2+y^2-2x+2y-47=0$$

D
$$x^2+y^2-2x-2y-62=0$$

Solution

The centre of the required circle lies at the intersection of $$2x - 3y- 5 = 0$$ and $$3x - 4y - 7 = 0$$.

Thus, the coordinates of the centre are $$(1, -1)$$.

Let $$r$$ be the radius of the circle.

$$\pi r^2 =154 $$

$$\Rightarrow \dfrac{22}{7}r^2=154 \Rightarrow r=7$$

Hence, the equation of required circle is $$(x -1)^2 +(y+ 1)^2= 7^2$$

$$\Rightarrow x^2+y^2-2x+2y-47=0$$
Question 3
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Let $$ABCD$$ be a square of side length $$1$$. and $$\Gamma $$ a circle passing through $$B$$ and $$C$$, and touching $$AD$$. The radius of $$\Gamma $$ is

A
$$\cfrac { 3 }{ 8 } $$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\cfrac { 1 }{ \sqrt { 2 } } $$

D
$$\cfrac { 5 }{ 8 } $$

Solution

$$PC=r$$
$${ PC }^{ 2 }={ r }^{ 2 }$$
$${ \left( r-1 \right) }^{ 2 }+{ \left( \cfrac { 1 }{ 2 } -1 \right) }^{ 2 }={ r }^{ 2 }$$
$$1-2r+\cfrac { 1 }{ 4 } =0$$
$$r=\cfrac { 5 }{ 8 } $$
Question 4
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Consider the parametric equation
$$x = \dfrac {a(1 - t^{2})}{1 + t^{2}}, y = \dfrac {2at}{1 + t^{2}}$$.
What does the equation represent?

A
It represents a circle of diameter $$a$$

B
It represents a circle of radius $$a$$

C
It represents a parabola

D
None of the above

Solution

$$x=\dfrac { a(1-{ t }^{ 2 }) }{ 1+{ t }^{ 2 } }, y=\dfrac { 2at }{ 1+{ t }^{ 2 } }$$
Let $$t=\tan\theta$$
$$\Rightarrow x=a\sin 2\theta , y=a\cos 2\theta \\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$
It represents a circle of radius $$a$$.
Question 5
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What is the radius of the circle passing through the point $$(2, 4)$$ and having centre at the intersection of the lines $$x - y = 4$$ and $$2x + 3y + 7 = 0$$?

A
$$3$$ units

B
$$5$$ units

C
$$3 \sqrt 3$$ units

D
$$5 \sqrt 2$$ units

Solution

The line $$x-y=4$$ and $$2x+3y+7=0$$ intersect at the point $$(1,-3)$$
So, the centre of circle lies at $$(1,-3)$$

Point $$(2,4)$$ lies on the circle.
So, radius = distance of $$(2,4)$$ from $$(1,-3)$$
$$=\sqrt{(2-1)^2+(4-(-3)^2}$$
$$=\sqrt{1^2+7^2}$$
$$=\sqrt{50}$$
$$=5\sqrt2$$ units

So, the answer is option (D)
Question 6
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The differential equation $$(3x + 4y + 1)dx + (4x + 5y + 1) dy = 0$$ represents a family of

A
Circles

B
Parabolas

C
Ellipses

D
Hyperbolas

Solution

The given differential equation is
$$(3x + 4y + 1)dx + (4x + 5y + 1)dy = 0... (i)$$
Comparing eq. (i) with $$Mdx + Ndy = 0$$,
we get
$$M = 3x + 4y + 1$$
and $$N = 4x + 5y + 1$$
Here, $$\dfrac {\partial M}{\partial y} = \dfrac {\partial N}{\partial x} = 4$$
Hence, eq. (i) is exact and solution is given by
$$\int (3x + 4y + 1)dx + \int (4x + 5y + 1)dy = C$$
$$\Rightarrow \dfrac {3x^{2}}{2} + 4xy + x + 4xy + \dfrac {5y^{2}}{2} + y - C = 0$$
$$\Rightarrow 3x^{2} + 16xy + 2x + 5y^{2} + 2y - 2C = 0$$
$$\Rightarrow 3x^{2} + 2\times 8xy + 2x + 5y^{2} + 2y + C' = 0 ... (ii)$$
where, $$C' = -2C$$
On comparing eq. (ii) with standard form of conic section
$$ax^{2} + 2hxy + by^{2} + 2gx + 2fy + C = 0$$
We get,
$$a = 3, h = 8, b = 5$$
Here, $$h^{2} - ab = 64 - 15 = 49 > 0$$
Hence, the solution of differential equation represents family of hyperbolas.
Question 7
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The line $$(x-2)\cos \theta +(y-2)\sin \theta =1$$ touches a circle for all value of $$\theta$$, then the equation of circle is

A
$$x^2+y^2-4x-4y+7=0$$

B
$$x^2+y^2+4x+4y+7=0$$

C
$$x^2+y^2-4x-4y-7=0$$

D
$$None\ of\ the\ above$$

Solution

Given line is
$$(x-2)cos \,\theta +(y-2)sin\, \theta =1$$
$$\Rightarrow (x-2)cos \theta +(y-2)sin\theta=cos^2 \, \theta+sin^2\, \theta$$
On compaining we get,
$$(x-2) = cos\theta $$ ..... $$(i)$$
$$(y-2) = sin \theta$$ ..... $$(ii)$$
On squaring and then adding Eqs. $$(i)$$ and $$(ii),$$ we get
$$(x-2)^2+(y-2)^2=cos^2\theta +sin^2 \theta$$
$$\Rightarrow (x-2)^2+(y-2)^2=1$$
$$\Rightarrow x^2+y^2-4x-4y+7=0$$
Question 8
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The latus rectum of the ellipse is half the minor axis. Then its eccentricity is

A
$$\dfrac {1}{\sqrt {2}}$$

B
$$\dfrac {1}{\sqrt {3}}$$

C
$$\dfrac {\sqrt {3}}{2}$$

D
None of these

Solution

$$\textbf{Step-1: Finding relation between a and b}$$
$$\text{Latus rectum of ellipse is half of the minor axis}$$
$$\Rightarrow \text{L.R} = \dfrac{{2{b^2}}}{a} = {b}$$
$$ \Rightarrow \dfrac{b}{a} = \dfrac{1}{2}$$ $$\dots(1)$$
$$\textbf{Step-2: Eccentricity of ellipse is given as-}$$
$$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} $$ $$\dots(2)$$
$$\textbf{Step-3: Putting values (1) in (2)}$$
$$e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}}$$
$$ = \sqrt {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} $$
$$= \sqrt {1 - \dfrac{1}{{4}}} $$
$$ = \sqrt {\dfrac{{3}}{{4}}}$$
$$\textbf{ So, the eccentricity of ellipse is given by (C)}$$ $$ \mathbf{{\dfrac{{\sqrt3}}{{2}}}}. $$
Question 9
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The equation of the smallest circle passing through the points $$(2, 2)$$ and $$(3, 3)$$ is

A
$$x^{2} + y^{2} + 5x + 5y + 12 = 0$$

B
$$x^{2} + y^{2} - 5x - 5y + 12 = 0$$

C
$$x^{2} + y^{2} + 5x - 5y + 12 = 0$$

D
$$x^{2} + y^{2} - 5x + 5y + 12 = 0$$

Solution

If the smallest circle is drawn through the points $$(2,2)$$ and $$(3,3)$$ then, the point have to be the opposite end of the diameter of the circle.
Therefore, the centre of the circle is $$C=(\dfrac{5}{2},\dfrac{5}{2})$$.
The radius of the circle is $$R=\dfrac{\sqrt{2}}{2}$$ ....(using distance formula)
Hence equation of the circle is
$$(x-\dfrac{5}{2})^{2}+(y-\dfrac{5}{2})^{2}=\dfrac{1}{2}$$
$$x^{2}+y^{2}-5x-5y+\dfrac{50}{4}=\dfrac{1}{2}$$
Or
$$x^{2}+y^{2}-5x-5y+\dfrac{25}{2}-\dfrac{1}{2}=0$$
Or
$$x^{2}+y^{2}-5x-5y+12=0$$.
Question 10
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If focii of $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ coincide with the focii of $$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$$ and eccentricity of the hyperbola is $$2$$, then

A
$$a^2+b^2=14$$

B
There is a director circle of the hyperbola

C
Centre of the director circle is $$(0,0)$$

D
Length of latus rectum of the hyperbola is $$12$$

Solution

For the ellipse, $$a=5, b=3$$ and $$e=\sqrt{\dfrac{25-9}{25}}=\dfrac{4}{5}$$
$$\therefore ae=4$$
Hence, the focii are $$(-4, 0)$$ and $$(4, 0)$$
As the focii of ellipse and hyperbola coincide
For the hyperbola, $$ae=4, e=2$$
$$\therefore a=2$$
$$b^2=a^{2}(e^{2}-1)=4(4-1)=12$$
$$\therefore b=\sqrt{12}$$
Length of latus rectum
$$=\dfrac{2 b^2}{a}=2 \times \dfrac{12}{2}$$
$$=12$$
Hence, option D is correct.

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Conic Sections Test - 25

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Conic Sections Test - 25

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SHARING IS CARING

Question 1

$$3{ x }^{ 2 }-{ y }^{ 2 }=48$$

$$4{ x }^{ 2 }-{ y }^{ 2 }=48$$

$${ x }^{ 2 }-3{ y }^{ 2 }=48$$

$${ x }^{ 2 }-4{ y }^{ 2 }=48$$

Solution

Question 2

$$x^2+y^2+2x-2y-62=0$$

$$x^2+y^2+2x-2y-47=0$$

$$x^2+y^2-2x+2y-47=0$$

$$x^2+y^2-2x-2y-62=0$$

Solution

Question 3

$$\cfrac { 3 }{ 8 } $$

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 1 }{ \sqrt { 2 } } $$

$$\cfrac { 5 }{ 8 } $$

Solution

Question 4

It represents a circle of diameter $$a$$

It represents a circle of radius $$a$$

It represents a parabola

None of the above

Solution

Question 5

$$3$$ units

$$5$$ units

$$3 \sqrt 3$$ units

$$5 \sqrt 2$$ units

Solution

Question 6

Circles

Parabolas

Ellipses

Hyperbolas

Solution

Question 7

$$x^2+y^2-4x-4y+7=0$$

$$x^2+y^2+4x+4y+7=0$$

$$x^2+y^2-4x-4y-7=0$$

$$None\ of\ the\ above$$

Solution

Question 8

$$\dfrac {1}{\sqrt {2}}$$

$$\dfrac {1}{\sqrt {3}}$$

$$\dfrac {\sqrt {3}}{2}$$

None of these

Solution

Question 9

$$x^{2} + y^{2} + 5x + 5y + 12 = 0$$

$$x^{2} + y^{2} - 5x - 5y + 12 = 0$$

$$x^{2} + y^{2} + 5x - 5y + 12 = 0$$

$$x^{2} + y^{2} - 5x + 5y + 12 = 0$$

Solution

Question 10

$$a^2+b^2=14$$

There is a director circle of the hyperbola

Centre of the director circle is $$(0,0)$$

Length of latus rectum of the hyperbola is $$12$$

Solution

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