Conic Sections Test

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Conic Sections Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

An ellipse has $$OB$$ as semi-minor axis, $$F$$ and $${ F }^{ ' }$$ its foci and the $$\angle FB{ F }^{ ' }$$ is a right angle. Then, the eccentricity of the ellipse is

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

B
$$\dfrac { 1 }{ 4 } $$

C
$$\dfrac { 1 }{ 2 } $$

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

Solution

Given, $$F$$ and $${ F }^{ ' }$$ are foci of an ellipse, whose coordinates are $$\left( ae,0 \right) $$ and $$\left( -ae,0 \right) $$ respectively and coordinates of $$B$$ are $$\left( 0,b \right) $$.
Slope of $$BF=\dfrac { b }{ -ae } $$
and slope of $$B{ F }^{ ' }=\dfrac { b }{ ae } $$
Since, $$ \angle FB{ F }^{ ' }={ 90 }^{ o }$$
Thus $$-\dfrac { b }{ ae } \left( \dfrac { b }{ ae } \right) =-1$$
$$\Rightarrow { b }^{ 2 }={ a }^{ 2 }{ e }^{ 2 }$$
Here $$ { e }^{ 2 }=1-\dfrac { { b }^{ 2 } }{ { a }^{ 2 } } =1-\dfrac { { a }^{ 2 }{ e }^{ 2 } }{ { a }^{ 2 } } $$
$$\Rightarrow 2{ e }^{ 2 }=1$$
$$\Rightarrow e=\dfrac { 1 }{ \sqrt { 2 } } $$
Question 2
1 / -0

If $$ e$$ and $$e'$$ be the eccentricities of a hyperbola and its conjugate, them $$ \dfrac {1}{e^2} + \dfrac {1}{e'^{2}} $$ is equal to :

A
0

B
1

C
2

D
None of these

Solution

$$ e^2 = \dfrac {a^2 + b^2}{a^2} $$ and $$ e'^{2} = \dfrac {a^2 + b^2}{b^2} $$

$$ \Rightarrow \dfrac {1}{e^2} + \dfrac {1}{e'^2} = \dfrac {a^2}{a^2+b^2} + \dfrac {b^2}{a^2 + b^2} = 1 $$
Question 3
1 / -0

The equation of the circle whose two diameters are the lines $$x+y=4$$ and $$x-y=2$$ and which passes through $$(4 , 6 )$$ is

A
$$x^2+y^2-6x-2y-16=0$$

B
$$x^2+y^2-6x-2y=15$$

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

D
$$5(x^2+y^2)-4x=16$$

Solution

Two diameters are the lines,
$$x+y=4$$ ----- ( 1 )
$$x-y=2$$ ----- ( 2 )

So, first, we find the intersection point of diameters that is the center of the circle.
Adding equation ( 1 ) and ( 2 ) we get,
$$\Rightarrow$$ $$2x=6$$
$$\Rightarrow$$ $$x=3$$

Substituting value of $$x$$ in equation ( 1 ),
$$\Rightarrow$$ $$3+y=4$$
$$\Rightarrow$$ $$y=1$$

So, the center of circle is $$(3,1).$$
Now, circle is pass through $$(4,6)$$
The equation of circle is $$(x-a)^2+(y-b)=r^2$$, where $$(a,b)$$ co-ordinates of circle and $$r$$ is radius.

Put center coordinates and pass through point
$$\Rightarrow$$ $$(4-3)^2+(6-1)^2=r^2$$
$$\Rightarrow$$ $$1+25=r^2$$
$$\Rightarrow$$ $$r^2=26$$

So, the equation of circle is,
$$\Rightarrow$$ $$(x-3)^2+(y-1)^2=26$$
$$\Rightarrow$$ $$x^2-6x+9+y^2-2y+1=26$$
$$\Rightarrow$$ $$x^2+y^2-6x-2y-16=0$$
Question 4
1 / -0

The equation of the circle passing through $$(2,0)$$ and $$(0,4)$$ and having the minimum radius is

A
$$x^{2}+y^{2}=20$$

B
$$x^{2}+y^{2}-2x-4y=0$$

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

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

Solution

Given points are $$(2,0)$$ and $$(0,4)$$

Therefore, equation of circle is $$(x-2)(x-0)+(y-0)(y-4)=0$$

By expanding, we get

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

Option B is correct.
Question 5
1 / -0

If the eccentricities of the hyperbola $$\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} = 1$$ and $$\dfrac {y^{2}}{b^{2}} - \dfrac {x^{2}}{a^{2}} = 1$$ be $$e$$ and $$e_{1}$$, then $$\dfrac {1}{e^{2}} + \dfrac {1}{e_{1}^{2}} =$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
None of these

Solution

Eccentricity of hyperbola is always > 1
Hence $$\dfrac{1}{e} < 1$$ so $$\dfrac{1}{e}+\dfrac{1}{e_1} <2$$

Eccentricity of $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}=1$$ is $$e=\sqrt{1+\dfrac{b^2}{a^2}}$$
In case of the second hyperbola the axes gets interchanged and eccentricity is defined as the deviation of the curve from being circular.
$$\implies$$ $$e_1=\sqrt{1+\dfrac{a^2}{b^2}}$$

$$\implies$$ $$\dfrac{1}{e^2} + \dfrac{1}{e_1^2} = $$ $$\dfrac{a^2}{a^2+b^2} + \dfrac{b^2}{a^2+b^2}=1$$
Question 6
1 / -0

The equation of the latus rectum of the parabola $$x^{2} + 4x + 2y = 0$$ is

A
$$2y + 3 = 0$$

B
$$3y = 2$$

C
$$2y = 3$$

D
$$3y + 2 = 0$$

Solution

$$x^2+4x+2y=0\rightarrow x^2+4x+4+2y-4=0\rightarrow (x+2)^2=2(2-y)$$
Now, focus of parabola $$(x-h)^2=4p(y-k)$$ is $$F(h,k+p)$$
Here $$h=-2,p=-\dfrac{1}{2},k=2\rightarrow F(-2,\dfrac{3}{2})$$
Equation of axis is $$x-h=0\rightarrow x+2=0$$
Now latus rectum is perpendicular to axis and passes through $$F$$.
$$\therefore$$Slope of latus rectum$$=0$$
Equation will be $$y-\dfrac{3}{2}=0(x-(-2))\rightarrow y=\dfrac{3}{2}\rightarrow 2y=3$$
Hence, $$(C)$$.
Question 7
1 / -0

Find the equation of the circle which touches the coordinate axes and whose centre lies on the line $$x - 2y = 3$$.

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

B
$$(x+1)^2+(y+1)^2=1$$

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

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

Solution

Given:
The equation of line is $$ x - 2y = 3 $$
Substitute $$x = 0$$ in the given equation.

$$ 0 - 2y = 3 \Rightarrow - 2y = 3 \Rightarrow y = - \dfrac{3}{2}$$

Substitute $$y = 0$$ in the given equation.

$$ x - 2\left (0 \right ) = 3 \Rightarrow x = 3 $$

The points are $$ \left (3, 0 \right ) \, and \, \left (0, - \dfrac{3}{2} \right ) $$.

Let the centre be $$ \left (a, -a \right ) $$.

Substitute $$ \left (a, -a \right ) $$ in given equation.

$$ a + 2a = 3 \Rightarrow 3a = 3 \Rightarrow a = 1 $$

Centre $$ = \left (1, -1 \right ) $$

Radius $$ = 1$$

The equation of a circle becomes $$ \left (x - 1 \right )^{2} + \left (y + 1 \right )^{2} = 1 $$.
Hence, the required solution is found.
Question 8
1 / -0

Consider a rigid square $$ABCD$$ as in the figure with $$A$$ and $$B$$ on the $$x$$ and $$y$$ axis respectively. When $$A$$ and $$B$$ slide along their respective axes, the locus of $$C$$ forms a part of.

A
A circle

B
A parabola

C
A hyperbola

D
An ellipse which is not a circle

Solution

Let the coordinates of C be (h,k). So, the coordinates of A and B can be determined as shown. The locus of C will be an equation containing the terms h and k.
Now, length of AB is constant. Using the distance formula, we get:
$$(0-h)^2 + ( \frac{k}{2} - 0)^2 = constant \implies h^2 + \frac{k^2}{4} = constant$$
We see that this equation is similar to that of an ellipse, which is
$$\frac{x^2}{a^2}+ \frac{y^2}{b^2} = constant$$
So, we can conclude that the locus of C is an ellipse. Here, $$a=1$$ and $$b=2$$.
Since $$a \neq b$$, it is not a circle.
Question 9
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The equation of the image of the circle $${ x }^{ 2 }+{ y }^{ 2 }+16x-24y+183=0$$ by the line mirror $$4x+7y+13=0$$ is:

A
$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y+235=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y-235=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y-235=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$$

Solution

Equation of given circle:
$$x^2+y^2++16x-24y+183=0$$

$$\textbf{Idea}$$; Lets find the reflection of centre of this circle with respect to the given line equation. Then find the radius of given circle. After reflection also, the radius of circle does not change. Thus finally knowing the centre of reflected circle and its radius we can find equation of reflected circle.

First convert this equation to standard form of circle equation as follows:
$$x^2+y^2++16x-24y+183=0$$
$$(x^2+16x+64)+(y^2-24y+144)-25=0$$
$$(x+8)^2+(y-12)^2-25=0$$
$$(x+8)^2+(y-12)^2=5^2$$
Thus $$\text{Centre}= (-8,12)$$
and $$\text{radius}=\sqrt{25}=5$$

Let reflected centre have co-ordinates= $$(h,k)$$
Now note that mid point of actual centre and reflected centre will definitely fall on line of reflection.
Mid-point of $$(h,k)$$ and $$(-8,12)$$=$$\left (\dfrac{h-8}{2},\dfrac{k+12}{2}\right)$$
Lets try to satisfy line equation with given point.
$$4x+7y+13=0\rightarrow4\left (\dfrac{h-8}{2}\right)+7\left (\dfrac{k+12}{2}\right)+13=0$$
$$4h+7k+78=0$$ .....(1)

Also slope of given line is perpendicular to slope of RC
Slope of given line$$=-4/7$$
$$m_1 m_2=-1$$
$$\dfrac{\dfrac{(k+12)}{2}-12} { \left (\dfrac{h-8}{2}\right) +8}\times\dfrac{-4}{7}=-1$$
$$\rightarrow7h-4k+104=0$$ ...........(2)
Solving (1) and (2) we get, $$(h,k)=(-16,-2)$$
Equation of circle with centre $$(-16,-2)$$ and radius $$=5$$:
$$(x+16)^2+(y+2)^2=5^2$$
$$x^2+y^2+32x+4y+235=0$$
Question 10
1 / -0

Suppose the parabola $$(y - k)^{2} = 4(x - h)$$ with vertex $$A$$, passes through $$O = (0, 0)$$ and $$L = (0, 2)$$. Let $$D$$ be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at $$P$$. Then $$\angle PDA$$ is equal to

A
$$\tan^{-1} \dfrac {1}{19}$$

B
$$\tan^{-1} \dfrac {2}{19}$$

C
$$\tan^{-1} \dfrac {4}{19}$$

D
$$\tan^{-1} \dfrac {8}{19}$$

Solution

Given Parabola: $$({ y-k) }^{ 2 }=4(x-h)\quad .............(i)$$
Vertex $$A$$:$$(h,k)$$
Focus $$S$$:$$(1+h,k)$$
Passes through $$O(0,0)$$ $$=>({ 0+k })^{ 2 }=4(0-h)$$
$$=>{ k }^{ 2 }=-4h\quad ...............(ii)$$
Also Passes through $$\angle (0,2)=>({ 2-k })^{ 2 }=4(0-h)\quad .................(iii)$$
From $$(ii)$$and$$(iii)$$, $${ k }^{ 2 }={ k }^{ 2 }+4-4k$$
$$=>k=1$$ and $$h=\cfrac { -1 }{ 4 } ...................(iv)$$
End point of $$\angle R:(1-\cfrac { 1 }{ 4 } ,1\pm 2),\quad D:(\cfrac { 3 }{ 4 } ,3)\quad and\quad (\cfrac { 3 }{ 4 } ,-1)$$
Axis of parabola: $$y-k=0 =>y=1............(v)$$
Point of intersection of $$y=1$$ with $$y-axis$$ ie. $$x=0$$ is $$P:(0,1)$$
Line Passing through $$PD$$:$$(y-1)=(\cfrac { 3-1 }{ \cfrac { 3 }{ 4 } } )(x-0)$$
$$:y=\cfrac { 8 }{ 3 } x+1$$
Slope of $$PD$$, $$m=\cfrac { 8 }{ 3 } ......................(vi)$$
Line Passing through $$AD$$, $$(y-1)=(\cfrac { 3-1 }{ \cfrac { 3 }{ 4 } +\cfrac { 1 }{ 4 } } )(x+\cfrac { 1 }{ 4 } )$$
$$=>y= 2x+\cfrac { 1 }{ 2 } +1$$
Slope of $$AD= { m }_{ 2 }=2..........(vii)$$
Angle between $$AD$$ and $$PD=>\angle PDA={ tan }^{ -1 }\left| \cfrac { \cfrac { 8 }{ 3 } -2 }{ 1+\cfrac { 8 }{ 3 } (2) } \right| $$
$$\angle PDA={ tan }^{ -1 }\left| \cfrac { \cfrac { 2 }{ 3 } }{ \cfrac { 19 }{ 3 } } \right| $$
$$\angle PDA={ tan }^{ -1 }(\cfrac { 2 }{ 19 } )$$.

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 28

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Conic Sections Test - 28

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SHARING IS CARING

Question 1

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

$$\dfrac { 1 }{ 4 } $$

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

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

Solution

Question 2

0

1

2

None of these

Solution

Question 3

$$x^2+y^2-6x-2y-16=0$$

$$x^2+y^2-6x-2y=15$$

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

$$5(x^2+y^2)-4x=16$$

Solution

Question 4

$$x^{2}+y^{2}=20$$

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

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

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

Solution

Question 5

$$1$$

$$2$$

$$3$$

None of these

Solution

Question 6

$$2y + 3 = 0$$

$$3y = 2$$

$$2y = 3$$

$$3y + 2 = 0$$

Solution

Question 7

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

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

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

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

Solution

Question 8

A circle

A parabola

A hyperbola

An ellipse which is not a circle

Solution

Question 9

$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y+235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y-235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x-4y-235=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$$

Solution

Question 10

$$\tan^{-1} \dfrac {1}{19}$$

$$\tan^{-1} \dfrac {2}{19}$$

$$\tan^{-1} \dfrac {4}{19}$$

$$\tan^{-1} \dfrac {8}{19}$$

Solution

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