Conic Sections Test

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Conic Sections Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

$$Ay^{2} + By + Cx + D = 0, (ABC\neq 0)$$ be the equation of a parabola, then

A
Then length of the latus rectum is $$\left |\dfrac {C}{A}\right |$$

B
The axis of the parabola is vertical

C
The y-coordinate of the vertex is $$-\dfrac {B}{2A}$$

D
The x-coordinate of the vertex is $$\dfrac {D}{A} + \dfrac {B^{2}}{4AC}$$

Solution

When, $$x={ A }\prime { y }^{ 2 }+B\prime { y }^{ 2 }+C\prime \quad $$ then length of $$LR$$ is $$\cfrac { 1 }{ \left| A\prime \right| } \quad ..................(1)$$
Given: $$A{ y }^{ 2 }+By+Cx+D=0$$
Converting the given equation into already known from ie. equation $$(1)$$,
$$=>A{ y }^{ 2 }+By+D=-Cx$$
$$=>x=\cfrac { A }{ C } { y }^{ 2 }+\cfrac { B }{ -C } y+\cfrac { D }{ -C } \quad ...................(2)$$
Comparing equation $$(2)$$ with equation $$(1)$$,
$$A\prime =\cfrac { A }{ -C } $$
Now length of $$LR=\cfrac { 1 }{ \left| A\prime \right| } =\cfrac { 1 }{ \left| \cfrac { A }{ -C } \right| } =\cfrac { C }{ A } $$
Question 2
1 / -0

If the lines $$3x - 4y - 7 = 0$$ and $$2s - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi$$ square units, the equation of the circle is-

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

B
$$x^{2} + y^{2} - 2x + 2y - 62 = 0$$

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

D
$$x^{2} + y^{2} + 2x - 2y - 47 = 0$$

Solution

Given: Area of circle=$$49\pi \ sq.units$$

$$\therefore$$ Area of circle=$$\pi \ r^2= 49\pi \ sq.units$$
where, $$r$$ is the radius of the circle
$$\pi \ r^2= 49\pi$$
$$r^2= 49$$
$$r= 7 \ units$$

Given eqs of lines are, $$3x-4y-7=0$$ and $$2x-3y-5=0$$

Solving the equations, we get $$x=1 \ , \ y=-1$$

The points of intersection of the lines is the center of the circle i.e., $$(1,-1)$$

We have the eqn for the circle with center $$(a,b)$$ and radius $$r$$,

$$(x-a)^2+(y-b)^2=(r)^2$$
$$(x-1)^2+(y+1)^2=(7)^2$$
$$(x^2+1-2x)+(y^2+1+2y)=49$$
$$x^2+y^2+2-2x+2y=49$$
$$x^2+y^2-2x+2y=49-2$$

$$\therefore$$ The equation of the given circle is,

$$x^2+y^2-2x+2y-47=0$$
Question 3
1 / -0

If the locus of the point $$(4t^2 - 1, 8t-2)$$ represents a parabola then the equation of latus rectum is

A
$$x - 5 = 0$$

B
$$2x - 7 = 0$$

C
$$x + 5 = 0$$

D
$$x -3 = 0$$

Solution

$$x=4t^2-1,y=8t-2$$

$$x+1=4t^2,y+2=8t$$

let $$X=x+1;Y=y+2$$

then the new coordinates will be $$X=4t^2, Y=8t$$

$$Y^2=4aX$$ is the general equation of a parabola

$$(8t)^2=4a\times 4t^2$$

$$64t^2=16at^2\Rightarrow a=4$$

Equation of latus rectum of $$Y^2=4aX$$ is $$X=a$$

But we know that $$X=x+1$$ ,substituting that in $$X=a$$

$$x+1=4\Rightarrow x-3=0$$
Question 4
1 / -0

If $$A(5,-4)$$ and $$B(7,6)$$ are points in a plane, then the set of all points $$P(x,y)$$ in the plane such that $$AP=PB=2:3$$ is

A
a circle

B
a hyperbola

C
an ellipse

D
a parabola
Question 5
1 / -0

The equation of the circle which passes through the points $$(2, 3)$$ and $$(4. 5)$$ and the centre lies on the straight line $$y - 4x + 3 = 0$$, is

A
$$x^{2} + y^{2} + 4x - 10y + 25 = 0$$

B
$$x^{2} - y^{2} - 4x - 10y + 25 = 0$$

C
$$x^{2} + y^{2} - 4x - 10y + 16 = 0$$

D
$$x^{2} + y^{2} - 14y + 8 = 0$$

Solution

Centre lies on the line $$y-4x+3=0$$

Let $$x=h$$

$$\Rightarrow y=4h-3$$

So the center is of the form $$(h,4h-3)$$

Distance of centre from $$(2,3)$$ and $$(4,5)$$ will be equal

$$\Rightarrow { (h-2) }^{ 2 }+{ (4h-3-3) }^{ 2 }={ (h-4) }^{ 2 }+{ (4h-3-5) }^{ 2 }\\ \Rightarrow h=2$$

So the centre is $$(2,5)$$

$$r=\sqrt { { (2-2) }^{ 2 }+{ (5-3) }^{ 2 } } =2$$

So the equation of the circle is

$${ (x-2) }^{ 2 }+{ (y-5) }^{ 2 }={ 2 }^{ 2 }\\ { x }^{ 2 }+{ y }^{ 2 }-4x-10y+25=0$$
Question 6
1 / -0

The equation of the circle, which is the mirror image of the circle, $${ x }^{ 2 }+{ y }^{ 2 }-2x=0$$, in the line, $$y=3-x$$ is:

A
$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+24=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+24=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad $$

D
$${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12=0\quad $$

Solution

Centre of circle $$(1,0)$$, radius $$= 1$$
Image of $$(1,0)$$ w.r.t. Line $$x+y-3=0$$ is
$$\dfrac{x-1}{1} = \dfrac{y-0}{1} = \dfrac{-2(1+0-3)}{1+1}=2$$
$$\Rightarrow x=3, \,\,y=2$$
Now circle with centre $$(3,2)$$ and radius $$1$$ is $$(x-3)^2+(y-2)^2=1$$
$$\Rightarrow x^2+9-6x+y^2+4-4y=1$$
$$\Rightarrow x^2+y^2-6x-4y+12=0$$
Question 7
1 / -0

Point $$\left( 0,\lambda \right) $$ lies in the interior of circle $$x^2+y^2=c^2$$ then

A
$$\lambda \in \left( -c,c\right) $$

B
$$\lambda \in \left(0,c \right) $$

C
$$\lambda \in \left(-c,0\right) $$

D
None.

Solution

The given equation is $$x^2+y^2=c^2$$
Condition of a point to be interior of circle
$$\implies x^2+y^2<c^2\\\implies 0+\lambda ^2<c^2\\\lambda ^2<c^2\\\lambda\in(-c,c)$$
Question 8
1 / -0

The equation of the parabola with vertex at (0, 0), axis along x-axis and passing through $$\displaystyle \left( \frac{5}{3}, \frac{10}{3} \right)$$ is

A
$$y^2 = 20 x$$

B
$$\displaystyle y^2 = \frac{20 x}{3}$$

C
$$\displaystyle y^2 = \frac{10 x}{3}$$

D
$$\displaystyle x^2 = \frac{20 y}{3}$$

Solution

The equation of parabola whose vertex is
at $$(0,0)$$ and the axis is along x axis
$$ y^{2} = \pm 4ax $$
as the curve is going through The point
$$ (5/3,10/3) $$
So, the curve is $$ y^{2} = 4ax $$
$$ \therefore (10/3)^{2} = 4.a(5/3) $$
$$ 100 = 4.a.15 $$
$$ a = 5/3 $$
$$ \therefore $$ The curve is $$ y^{2} = 4.5/3.x $$
$$ \therefore 3y^{2} = 20x $$ [Ans]
Question 9
1 / -0

If $$b$$ and $$c$$ are the lengths of the segments of any focal chord of a parabola $$y^{2} = 4ax$$, then length of the semi-latus rectum is

A
$$\dfrac{b+c}{2}$$

B
$$\dfrac{bc}{b+c}$$

C
$$\dfrac{2bc}{b+c}$$

D
$$\sqrt{bc}$$

Solution

We know that, the semi latus-rectum of a parabola is the harmonic mean between the segments of any focal chord of the parabola.

Given that, $$b$$ and $$c$$ are lengths of the segments of any focal chord of the parabola $${y}^{2}=4ax$$

So, Let $$l$$ is the length of semi-latus rectum of the parabola.

Then,$$\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{l}$$

$$\Rightarrow\,l=\dfrac{2bc}{b+c}$$

Hence, length of semi-latus-rectum$$=\dfrac{2bc}{b+c}$$
Question 10
1 / -0

State whether following statements are true or false
Statement-1 : The only circle having radius $$\sqrt {10}$$ and a diameter along line $$2x + y=5$$ is $$x^2 + y^2 - 6x + 2y=0$$.
Statement-2: The line 2x + y=5 is a normal to the circle $$x^2+ y^2 - 6x + 2y =0.$$

A
Statement- 1 is false, statement-2 is true.

B
Statement-1 is true,statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

C
Statement-1 is true, statement-2 is false.

D
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Solution

Statement 1 and 2 both are correct, but statement 2 is not correct explaination for statement 1.

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

$$(-g,-f)=(-3,1), c=0$$

$$r^2=g^2+f^2-c$$

$$\therefore r^2=9+1-0=10$$

$$\Rightarrow r=\sqrt{10}$$

$$2x+y=5$$ is also a normal to the circle.

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 29

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Conic Sections Test - 29

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Question 1

Then length of the latus rectum is $$\left |\dfrac {C}{A}\right |$$

The axis of the parabola is vertical

The y-coordinate of the vertex is $$-\dfrac {B}{2A}$$

The x-coordinate of the vertex is $$\dfrac {D}{A} + \dfrac {B^{2}}{4AC}$$

Solution

Question 2

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

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

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

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

Solution

Question 3

$$x - 5 = 0$$

$$2x - 7 = 0$$

$$x + 5 = 0$$

$$x -3 = 0$$

Solution

Question 4

a circle

a hyperbola

an ellipse

a parabola

Question 5

$$x^{2} + y^{2} + 4x - 10y + 25 = 0$$

$$x^{2} - y^{2} - 4x - 10y + 25 = 0$$

$$x^{2} + y^{2} - 4x - 10y + 16 = 0$$

$$x^{2} + y^{2} - 14y + 8 = 0$$

Solution

Question 6

$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+24=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+24=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad $$

$${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12=0\quad $$

Solution

Question 7

$$\lambda \in \left( -c,c\right) $$

$$\lambda \in \left(0,c \right) $$

$$\lambda \in \left(-c,0\right) $$

None.

Solution

Question 8

$$y^2 = 20 x$$

$$\displaystyle y^2 = \frac{20 x}{3}$$

$$\displaystyle y^2 = \frac{10 x}{3}$$

$$\displaystyle x^2 = \frac{20 y}{3}$$

Solution

Question 9

$$\dfrac{b+c}{2}$$

$$\dfrac{bc}{b+c}$$

$$\dfrac{2bc}{b+c}$$

$$\sqrt{bc}$$

Solution

Question 10

Statement- 1 is false, statement-2 is true.

Statement-1 is true,statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is false.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Solution

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