Conic Sections Test

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Conic Sections Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

A circle is concentric with circle $$x^{2}+ y^{2}-2x+4y-20=0$$. If perimeter of the semicircle is $$36$$ then the equation of the circle is :

A
$$x^{2}+y^{2}-2x-4y-44=0$$

B
$$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$$

C
$$x^{2}+y^{2}-2x+4y-43=0$$

D
$$none\ of\ these$$

Solution

$$x^2+y^2-2x+4y-20=0$$
center $$(1, -2)$$
perimeter $$\Rightarrow 36=(\pi r+2r)$$
$$36=r(3.14+2)=$$
$$((x-1)+y+2)^2=\left(\dfrac{126}{11}\right)^2$$ or $$(7)^2$$ $$r=\dfrac{36}{5.14}=7$$
Question 2
1 / -0

The axes are translated so that the new equation of the circle $$x^{2}+y^{2}-5x+2y-5=0$$ has no first degree terms. Then the new equation is

A
$$x^{2}+y^{2}=9$$

B
$$x^{2}+y^{2}=\dfrac {49}{4}$$

C
$$x^{2}+y^{2}=\dfrac {81}{16}$$

D
$$none\ of\ these$$

Solution

$$x^2+y^2-5x+2y-5=0$$
$$\sqrt {\dfrac {25}{4}+1+25}$$
$$=\sqrt {\dfrac {49}{4}}$$
$$ (5/2, -1)$$
$$x^2+y^2=\left (\sqrt {\dfrac {49}{4}}\right)^2$$
$$x^2+y^2=\dfrac {49}{4}$$
Question 3
1 / -0

vertices of an ellipse are $$(0,\pm 10)$$ and its eccentricity $$e=4/5$$ then its equation is

A
$$90x^2-40y^2=3600$$

B
$$80x^2+50y^2=4000$$

C
$$36x^2+100y^2=3600$$

D
$$100x^2+36y^2=3600$$

Solution

Let the equation of the required ellipse be
$$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\longrightarrow \left( 1 \right) $$
since the vertices of the ellipse are on $$y$$-axis, so the coordinate of the vertices are $$\left( 0,\pm b \right) $$
$$\therefore b=10\\ Now,\quad { a }^{ 2 }=b^{ 2 }\left( 1-{ e }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }=100\left( 1-\dfrac { 16 }{ 25 } \right) \\ \Rightarrow { a }^{ 2 }=36\\ $$
substituting the value of $${ a }^{ 2 }$$ and $${ b }^{ 2 }$$ in equation $$(1)$$
we get, $$\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$$
Question 4
1 / -0

The name of the conic represent by the equation $$x^2+y^2-2y+20x+10=0$$ is

A
a hyperbola

B
an ellipse

C
a parabola

D
circle

Solution

For a standard second degree equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$
to be a circle $$a=b\\h=0$$
Here $$a=b=1\\h=0$$
So The given equation is Circle
Question 5
1 / -0

The equation of the circle passing through the foci of the ellipes $${\frac{x}{{16}}^2} + {\frac{y}{{{9^{}}}}^2} = 1$$ and having centre at $$\left( {0,3} \right)$$ is

A
$${x^2} + {y^2} - 6y - 7 = 0$$

B
$${x^2} + {y^2} - 6y +7 = 0$$

C
$${x^2} + {y^2} - 6y - 5 = 0$$

D
$${x^2} + {y^2} - 6y + 5 = 0$$

Solution

R.E.F image
Equation of ellipse : $$ \frac{x^{2}}{16}+\frac{y^{2}}{9} = 1 $$
coordinate of foci are $$ (ac,o); (-ac,o) $$
$$ e = \sqrt{1-\frac{9}{16}} = \frac{\sqrt{7}}{4} $$
$$ a = 4 $$
co-ordinate of foci $$ = (\sqrt{7},0),(-\sqrt{7},0) $$
$$ R = \sqrt{7+9} = 4 $$
$$ (x-0)^{2}+(y-3)^{2} = 4^{2} $$
$$ x^{2}+y^{2}-6y+9 = 16 $$
$$ x^{2}+y^{2}-6y-7 = 0 $$
Question 6
1 / -0

The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is $$e=2/3$$

A
$$36x^2+20y^2=405$$

B
$$20x^2+36y^2=405$$

C
$$30x^2+22y^2=411$$

D
$$22x^2+32y^2=409$$

Solution

$$\begin{array}{l} e=\frac { 2 }{ 3 } =\sqrt { \frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ c } } \\ \Rightarrow { \left( { \frac { 2 }{ 3 } } \right) ^{ 2 } }=\frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ { { a^{ 2 } } } } \\ \Rightarrow \frac { { 4{ a^{ 2 } } } }{ a } ={ a^{ 2 } }-{ b^{ 2 } } \\ \Rightarrow { b^{ 2 } }={ a^{ 2 } }-4{ a^{ 2 } }=\frac { { 5{ a^{ 2 } } } }{ 9 } \to \left( i \right) \end{array}$$
Equation of Ellipse are
$$\begin{array}{l} \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { { y^{ 2 } } } }{ { { b^{ 2 } } } } =1 \\ \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { 9{ y^{ 2 } } } }{ { 5{ a^{ 2 } } } } =1\to \left( { ii } \right) \\ Put\, \, { a^{ 2 } }=\frac { { 405 } }{ { 20 } } \, \, \left( { From\, \, option\, \, in\, \, equation\left( i \right) } \right) \\ Then,\, \, { b^{ 2 } }=\frac { { 405 } }{ { 360 } } \end{array}$$
Hence, equation of ellipse is
$$ \Rightarrow \frac{{{x^2}}}{{\left( {\frac{{405}}{{20}}} \right)}} + \frac{{{y^2}}}{{\left( {\frac{{405}}{{36}}} \right)}} = - 1,20{x^2} + 36{y^2} = 405$$
Question 7
1 / -0

The equation $$\dfrac { x ^ { 2 } } { 10 - a } + \dfrac { y ^ { 2 } } { 4 - a } = 1$$ represents an ellipse if

A
$$a < 4$$

B
$$a > 4$$

C
$$4 < a < 10$$

D
None of these

Solution

$$\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1$$
For this equation to represent an ellipse its eccentricity shoule lie between $$0$$ and $$1$$.
$$\sqrt{1-\dfrac{b^2}{a^2}}< 1$$
$$0< 1-\dfrac{(4-a)^2}{(10-a)^2} <1$$
$$0 < (10-a)^2-(4-a)^2 <1$$
$$0< 84-12a <1$$
$$0< (7-a)12<1$$
$$12(7-a)> 0$$ and
$$12(7-a)<1$$
$$a< 7$$ and $$7-a<\dfrac{1}{12}$$
$$a< 7$$ and $$a >\dfrac{83}{12}$$
Question 8
1 / -0

Consider the set of hydperbola $$xy=k,k\ \in\ R$$. Let $$e_{1}$$ be the eccentricity when $$k=4$$ and $$e_{2}$$ be the eccentricity when $$k=9$$ . Then $$e^{2}_{1}+e^{2}_{2}=$$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution
Question 9
1 / -0

If a circle with centre $$(0,0)$$ touches the line $$5x+12y=1$$ then it equation will be

A
$$169(x^{2}+y^{2})=1$$

B
$$ (x^{2}+y^{2})=169$$

C
$$16(x^{2}+y^{2})=1$$

D
$$ (x^{2}+y^{2})=13$$

Solution
Question 10
1 / -0

If there is exactly one tangent at a distance of $$4$$ units from one of the locus of $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{a^{2}-16}=1, a>4$$, then length of latus rectum is :-

A
$$16$$

B
$$\dfrac{8}{3}$$

C
$$12$$

D
$$15$$

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 38

Result

Conic Sections Test - 38

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SHARING IS CARING

Question 1

$$x^{2}+y^{2}-2x-4y-44=0$$

$$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$$

$$x^{2}+y^{2}-2x+4y-43=0$$

$$none\ of\ these$$

Solution

Question 2

$$x^{2}+y^{2}=9$$

$$x^{2}+y^{2}=\dfrac {49}{4}$$

$$x^{2}+y^{2}=\dfrac {81}{16}$$

$$none\ of\ these$$

Solution

Question 3

$$90x^2-40y^2=3600$$

$$80x^2+50y^2=4000$$

$$36x^2+100y^2=3600$$

$$100x^2+36y^2=3600$$

Solution

Question 4

a hyperbola

an ellipse

a parabola

circle

Solution

Question 5

$${x^2} + {y^2} - 6y - 7 = 0$$

$${x^2} + {y^2} - 6y +7 = 0$$

$${x^2} + {y^2} - 6y - 5 = 0$$

$${x^2} + {y^2} - 6y + 5 = 0$$

Solution

Question 6

$$36x^2+20y^2=405$$

$$20x^2+36y^2=405$$

$$30x^2+22y^2=411$$

$$22x^2+32y^2=409$$

Solution

Question 7

$$a < 4$$

$$a > 4$$

$$4 < a < 10$$

None of these

Solution

Question 8

$$2$$

$$3$$

$$4$$

$$1$$

Solution

Question 9

$$169(x^{2}+y^{2})=1$$

$$ (x^{2}+y^{2})=169$$

$$16(x^{2}+y^{2})=1$$

$$ (x^{2}+y^{2})=13$$

Solution

Question 10

$$16$$

$$\dfrac{8}{3}$$

$$12$$

$$15$$

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