Conic Sections Test

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Conic Sections Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

What is the area enclosed by $$|x|+|y|=1$$ ?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$|x|+ |y|=1 \Rightarrow \begin{cases} x+y=1 & x \ge 0 , y \ge 0 \\ -x+y=1 & x < 0, y \ge 0 \\ -x-y=1 & x < 0 , y < 0 \\ x-y=1 & x \ge 0 , y < 0 \end{cases}$$
Plotting on graph
Area enclosed $$=1 (1) =1 unit^{2}$$
Question 2
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Equation of the circle which passes through the centre of the circle $$x^{2} + y^{2} + 8x + 10y - 7 = 0$$ and is concentric with the circle $$2x^{2} + 2y^{2} - 8x - 12y - 9 = 10$$ is

A
$$x^{2} + y^{2} - 4x - 8y - 97 = 0$$

B
$$x^{2} + y^{2} - 4x - 6y - 87 = 0$$

C
$$x^{2} + y^{2} - 2x - 8y - 95 = 0$$

D
None of these

Solution

First equation $$=x^2+y^2+8x+10y-7=0$$

its centre $$=(-4, -5)$$

Second equation of the circle

$$=2x^2+2y^2+8x-12y-9=0$$

its, centre $$=(2, 3)$$

radius:$$=r^2\sqrt {(2+4)^2+(3+5)^2}$$

$$r^2=\sqrt {6^2+8^2}=\sqrt {36+64}$$

$$r^2 =100$$

$$r=10$$

The equation of the concentric

circle $$=(x-2)^2+(y-3)^2=10^2$$

$$x^2-4x+4+y^2-6y+9-100=0$$

$$x^2-4x+y^2-6y+13-100=0$$

$$x^2+y^2-4x-6y-87=0$$
Question 3
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The centres of a set of circles, each of radius $$3$$, lie on the circle $${x}^{2}+{y}^{2}=25$$. The lotus of any point in the set is

A
$$4\le {x}^{2}+{y}^{2}\le 64$$

B
$${x}^{2}+{y}^{2}\le 25$$

C
$${x}^{2}+{y}^{2}\ge 25$$

D
$$3\le {x}^{2}+{y}^{2}\le 9$$

Solution
Question 4
1 / -0

The curve described parametrically by$$x = {t^2} + t + 1$$ and $$y = {t^2} - t + 1$$ represents

A
hyperbola

B
ellipse

C
parabola

D
rectangular hyperbola

Solution

The equations $$x=t^2+t+1$$ and $$y=t^2-t+1$$
Subtracting $$y$$ from $$x,$$ we get
$$x-y=2t$$
$$\Rightarrow$$ $$t=\dfrac{x-y}{2}$$ ----- ( 1 )
Now, adding $$x$$ and $$y$$ we get,
$$\Rightarrow$$ $$x+y=2t^2+2$$ ----- ( 2 )
Substituting the value of $$t$$ from equation ( 1 ) in equation ( 2 ), we get
$$x+y=2\left(\dfrac{x-y}{2}\right)^2+2$$

$$\Rightarrow$$ $$x+y=2\times\dfrac{x^2+y^2-2xy}{4}+2$$

$$\Rightarrow$$ $$x+y=\dfrac{x^2+y^2-2xy}{2}+2$$

$$\Rightarrow$$ $$2x+2y=x^2+y^2-2xy+4$$

$$\Rightarrow$$ $$x^2+y^2-2xy+4-2x-2y=0$$ ----- ( 3 )
The general second degree equation is of the form
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ ----- ( 4 )
Comparing equations ( 3 ) and ( 4 ), we have
$$a=1,h=-1,b=1,g=-1,f=-1$$ and $$c=4$$
This equation represents parabola if $$h^2-ab=0$$
$$h^2-ab=(-1)^2-(1)(1)=1-1=0$$
$$\therefore$$ The equation ( 3 ) represents a parabola.
Question 5
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If $$2x - 3y = 5$$ and $$3x - 4y = 7$$ are the equation of $$2$$ diameters of a circle whose area is $$88$$ sq. units then the equation of the circle is :

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

B
$${x^2} + {y^2} - 2x + 2y - 49 = 0$$

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

D
$${x^2} + {y^2} - 2x + 2y - 26 = 0$$

Solution
Question 6
1 / -0

If he equations of two diameters of a circle are $$2x + y = 6$$ and $$3x + 2y = 4$$ and the radius is $$10$$ , find the equation of the circle.

A
$$x^{2}+y^{2}-16x+20y+64=0$$

B
$$x^{2}-y^{2}+16x-20y+34=0$$

C
$$x^{2}+y^{2}+6x-20y+64=0$$

D
None of these

Solution

Equation of two diameters of circle are
$$2x+y=6 ....... (i)$$
$$3x+2y=4 ...... (ii)$$

We know intersection of these line will give centre of circle

$$4x+2y=12$$
$$3x+2y=4$$
$$(-)$$ $$(-)$$
$$\overline {2x+0=16}$$

$$\Rightarrow x=8$$

Put $$x=8$$ in $$(i)$$ we get $$16+y=6$$
$$y=-10$$

centre of circle is $$(8, -10)$$ and radius $$10$$ equation of circle will be

$$(x-8)^{2}+(y+10)^{2}=(10)^{2}$$

$$x^{2}-16x+64+y^{2}+20y+100=100$$

$$x^{2}+y^{2}-16x+20y+64=0$$
Question 7
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The equation $$\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$$ represents an ellipse if

A
$$r>1$$

B
$$r>5$$

C
$$2 < r< 5$$

D
$$r<2$$ or $$r>5$$

Solution
Question 8
1 / -0

The equation of directrix of a parabola $$3 x + 4 y + 15 = 0$$ and equation of tangent at vertex is $$3 x + 4 y - 5 = 0$$ . Then the length of latus rectum is equal to-

A
15

B
14

C
13

D
16

Solution
Question 9
1 / -0

Which of the following equations represents parametrically, parabolic profile ?

A
$$x=3\:cos\:t\:;\:y=4\:sin\:t$$

B
$$x^2-2=-\:cos\:t\:;\:y=4\:cos^2\:\dfrac{t}{2}$$

C
$$\sqrt{x}=tan\:t\:;\:\sqrt{y}=sec\:t$$

D
$$x=\sqrt{1-sin\:t};\:y=sin\:\dfrac{t}{2}+cos\:\dfrac{t}{2}$$

Solution

Option$$a)$$
$$x=3\cos{t}\Rightarrow \cos{t}=\dfrac{x}{3}$$

$$y=4\sin{t}\Rightarrow\sin{t}=\dfrac{y}{4}$$

We know the identity, $${\sin}^{2}{t}+{\cos}^{2}{t}=1$$

$$\Rightarrow {\left(\dfrac{y}{4}\right)}^{2}+{\left(\dfrac{x}{3}\right)}^{2}=1$$

$$\Rightarrow \dfrac{{y}^{2}}{16}+\dfrac{{x}^{2}}{9}=1$$

$$\Rightarrow \dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{16}=1$$ represents an equation of ellipse.

Option$$b)$$

$${x}^{2}-2=-2\cos{t}$$

$$\Rightarrow {x}^{2}=2-2\cos{t}$$

$$\Rightarrow {x}^{2}=2\left(1-\cos{t}\right)$$

$$\Rightarrow {x}^{2}=2\left(1-\left(1-2{\sin}^{2}{\dfrac{t}{2}}\right)\right)$$

$$\Rightarrow {x}^{2}=4{\sin}^{2}{\dfrac{t}{2}}$$

We have $$y=4{\cos}^{2}{\dfrac{t}{2}}$$

$$\Rightarrow {\cos}^{2}{\dfrac{t}{2}}=\dfrac{y}{4}$$

We know the identity, $${\sin}^{2}{\dfrac{t}{2}}+{\cos}^{2}{\dfrac{t}{2}}=1$$

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

$$\Rightarrow {x}^{2}=4-y$$ represents a parabolic profile.

Option$$c)$$
$$\sqrt{x}=\tan{t}, \sqrt{y}=\sec{t}$$

We know that $${\sec}^{2}{t}-{\tan}^{2}{t}=1$$

$$\Rightarrow{\left(\sqrt{y}\right)}^{2}-{\left(\sqrt{x}\right)}^{2}=1$$

$$\Rightarrow y-x=1$$ or $$x-y+1=0$$ is the equation of a straight line.

Option$$d)$$

$$x=\sqrt{1-\sin{t}}=\sqrt{{\sin}^{2}{t}+{\cos}^{2}{t}-\sin{t}}$$ using the identity $${\sin}^{2}{t}+{\cos}^{2}{t}=1$$

$$=\sqrt{{\sin}^{2}{t}+{\cos}^{2}{t}-2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}$$ using multiple angle formula $$\sin{t}=2\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}$$

$$=\sqrt{{\left(\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}\right)}^{2}}$$

$$=\left|\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}\right|$$

$$\therefore x=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}$$ and $$x=\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}$$

We have $$y=\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}$$

$$\Rightarrow x+y=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}+\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}=2\sin{\dfrac{t}{2}}$$

$$\Rightarrow \sin{\dfrac{t}{2}}=\dfrac{y+x}{2}$$

$$\Rightarrow x-y=\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}-\cos{\dfrac{t}{2}}=-2\cos{\dfrac{t}{2}}$$

$$\Rightarrow \cos{\dfrac{t}{2}}=\dfrac{y-x}{2}$$

We have the identity

$${\sin}^{2}{\dfrac{t}{2}}+{\cos}^{2}{\dfrac{t}{2}}=1$$

$$\Rightarrow {\left(\dfrac{y+x}{2}\right)}^{2}+{\left(\dfrac{y-x}{2}\right)}^{2}=1$$

$$\Rightarrow {x}^{2}+{y}^{2}+2xy+{x}^{2}+{y}^{2}-2xy=4$$

$$\Rightarrow {x}^{2}+{y}^{2}=2$$ is the equation of the circle.
Question 10
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Equation of the circle of radius 5 whose centre lies on y-axis in first quadrant and passes through$$\left( {3,\,\,\,\,2} \right)$$ is

A
$${x^2} + {y^2} - 12y + 11 = 0$$

B
$${x^2} + {y^2} - 6y - 1 = 0$$

C
$${x^2} + {y^2} - 8y + 3 = 0$$

D
None of these

Solution

Let the coordinate of the center be $$(0,k)$$ as center lies on y-axis

hence, the Equation of circle should be
$$(x-0)^2+(y-k)^2=5^2\Rightarrow x^2+y^2+k^2-2ky=25$$

and it passes through the point (3,2)
$$3^2+2^2+k^2-2k\cdot 2=25\Rightarrow k^2-4k-12=0\Rightarrow k=-2,6$$

But $$k=6$$ as the center is in the first quadrant

Putting this value in $$x^2+y^2+k^2-2ky=25$$, we get

$$x^2+y^2+36-12y=25\Rightarrow x^2+y^2-12y+11=0$$