Conic Sections Test

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Conic Sections Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The eccentricity of the hyperbola whose latus-return is $$8$$ and length of the conjugate axis is equal to half the distance between the foci, is

A
$$\dfrac43$$

B
$$\dfrac4{\surd 3}$$

C
$$\dfrac2{\surd 3}$$

D
$$None\ of\ these$$

Solution

Given that the length of the latus rectum is $$8$$ and length of the conjugate axis is equal to half the distance between the foci.

$$\Rightarrow \dfrac{2b^2}{a}=8$$ and $$2b=\dfrac{1}{2}(2ae)$$

$$\therefore \dfrac{2}{a}\left(\dfrac{ae}{2}\right)^2=8$$

$$\Rightarrow ae^2=16$$ ...(1)

We have $$\dfrac{2b^2}{a}=8$$

$$\Rightarrow b^2=4a$$

$$\Rightarrow a^2(e^2-1)=4a$$

$$\Rightarrow ae^2-a=4$$

$$\Rightarrow 16-a=4$$ (by (1))

$$\Rightarrow a=12$$

Substitute $$a=12$$ in (1)

$$\Rightarrow 12e^2=16$$

$$\Rightarrow e^2=\dfrac{4}{3}$$

$$\therefore e=\dfrac{2}{\sqrt{3}}$$
Question 2
1 / -0

The graph of curve
$${x^2} + {y^2} - 8x - 8y + 32 = 0$$ falls wholly in the

A
first quadrant

B
second quadrant

C
third quadrant

D
none of these

Solution

$${x}^{2}+{y}^{2}−2xy−8x−8y+32=0$$
$$\Rightarrow\,{x}^{2}+{y}^{2}−2xy=8x+8y-32$$
$$\Rightarrow {\left(x-y\right)}^{2}= 8\left(x + y - 4\right)$$
is a parabola whose axis is $$x - y = 0$$ and the tangent at the vertex is $$x + y - 4 = 0$$
Also, when $$y = 0$$, we have
$$x^2 - 8x + 32 = 0$$ which gives no real values of $$x$$
when$$ x = 0$$, we have $$y^2 - 8y + 32 = 0$$ which gives no real values of $$y$$.
So, the parabola does not intersect the axes. Hence, the graph falls in the first quadrant.
Question 3
1 / -0

If foci are points $$(0,1)(0,-1)$$ and minor axis is of length $$1$$, then equation of ellipse is

A
$$\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 5/4 } =1$$

B
$$\dfrac { { x }^{ 2 } }{ 5/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$$

C
$$\dfrac { { x }^{ 2 } }{ 3/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$$

D
$$\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 3/4 } =1$$

Solution

Given that focii are $$(0,1), (0,-1)$$

Axis lies along $$y-axis$$

Distance between focii$$\rightarrow 2be=2$$

$$\Rightarrow be=1$$

Given that $$2a=1$$

$$\Rightarrow a=\dfrac{1}{2}$$

We know that $$\Rightarrow e^2=1-\dfrac{a^2}{b^2}$$

$$\Rightarrow b^2e^2=b^2-a^2$$

Substituting above obtained values in this expression we get,

$$\Rightarrow 1=b^2-(\dfrac{1}{2})^2$$

$$\Rightarrow 1=b^2-\dfrac{1}{4}$$

$$\Rightarrow b^2=\dfrac{5}{4}$$

Thus equation of ellipse$$\Rightarrow \dfrac{x^2}{\dfrac{1}{4}}+\dfrac{y^2}{\dfrac{5}{4}}=1$$
Question 4
1 / -0

If a be the radius of a circle which touches x-axis at the origin, then its equation is

A
$${ x }^{ 2 }+{ y }^{ 2 }+ax=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }\pm 2ya=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }\pm 2xa=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }+ya=0$$

Solution

The equation of the circle with centre at $$\left(h,k\right)$$ and radius equal to $$a$$ is $${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$$
When the circle passes through the origin and centre lies on $$x-$$ axis
$$\Rightarrow h=a$$ and $$k=0$$
Then the equation $${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$$ becomes $${\left(x-a\right)}^{2}+{y}^{2}={a}^{2}$$
If a circle passes through the origin and centre lies on $$x-$$axis then the abscissa will be equal to the radius of the circle and the $$y-$$co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form
$${\left(x\pm a\right)}^{2}+{y}^{2}={a}^{2}$$
$$\Rightarrow {x}^{2}+{a}^{2}\pm 2ax+{y}^{2}={a}^{2}={x}^{2}+{y}^{2}\pm 2ax=0$$ is the required equation of the circle.
Question 5
1 / -0

The centre of the circle $$(x-a)(x-b)+(y-c)(y-c)=0$$ is

A
$$(a+b,c+d)$$

B
$$9-a-b,-c-d)$$

C
$$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$

D
none of these

Solution

Given the equation of the circle is
$$(x-a)(x-b)+(y-c)(y-c)=0$$
or, $$x^2-x(a+b)+ab+y^2-x(c+c)+cd=0$$
or, $$x^2-2.x.\dfrac{a+b}{2}+\dfrac{(a+b)^2}{4}+y^2-2.y.\dfrac{c+d}{2}+\dfrac{(c+d)^2}{4}=\dfrac{(a+b)^2}{4}-ab+\dfrac{(c+d)^2}{4}-cd$$
or, $$\left(x-\dfrac{a+b}{2}\right)^2+\left(y-\dfrac{c+d}{2}\right)^2=\dfrac{(a-b)^2}{4}+\dfrac{(c-d)^2}{4}$$.
From this equation it is clear that the centre is $$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$.
Question 6
1 / -0

If the vertex $$= (2, 0)$$ and the extremities of the latus rectum are $$(3, 2)$$ and $$(3, -2)$$, then the equation of the parabola is

A
$$y^2 = 2x - 4$$

B
$$x^2 = 4y - 8$$

C
$$y^2 = 4x - 8$$

D
None

Solution

$$y^2=4ax$$

$$(y-0)^2=4a(x-2)$$

$$y^2=4a(x-2)$$

$$y^2=4(x-2)$$

$$y^2=4x-8$$
Question 7
1 / -0

Equation of the circle with centre on the $$y-$$axis and passing through the origin and the point $$(2,3)$$ is

A
$$x^{2}+y^{2}+13y=0$$

B
$$3x^{2}+3y^{2}+13x+3=0$$

C
$$6x^{2}+6y^{2}-13y=0$$

D
$$x^{2}+y^{2}+13+3=0$$

Solution

We have to find equation of a circle with center on the y-axis.
General equation of such circle is
$$(x-0)^{2}+(y-k)^{2}=k^{2}$$
It passes through $$(2,3)$$
i.e, $$2^{2}+(3-k)^{2}=k^{2}$$
$$\Rightarrow 4+9+k^{2}-6k=k^{2}$$
$$\Rightarrow k=\dfrac{13}{6}$$
$$\therefore $$ Equation of circle $$=x^{2}+\left(y-\dfrac{13}{6}\right)^{2}=\left(\dfrac{13}{6}\right)^{2}$$
$$=6x^{2}+6y^{2}-13y=0$$
Question 8
1 / -0

The ellipse $$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$$ cuts x axis at A and y axis at B and the line joining the focus S and B makes an angle $$\dfrac{{3\pi }}{4}$$ with x-axis. Then the eccentricity of the ellipse is

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

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

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

D
$$\dfrac{1}{3}$$

Solution

REF.Image
B is (0,b) S is (ae,0)
$$\dfrac{-b}{ae}=tan\dfrac{3\pi }{4}$$
$$e=\dfrac{b}{a}$$
$$\dfrac{b}{a}=\sqrt{1-\dfrac{b^{2}}{a^{2}}}$$ $$\dfrac{b}{a}=\dfrac{1}{\sqrt{2}}$$
$$e=\dfrac{1}{\sqrt{2}}$$
A is correct
Question 9
1 / -0

If $$(4,3)$$ and $$(-12,-1)$$ are end points of a diameter of a circle, then the equation of the circle is-

A
$${ x }^{ 2 }+{ y }^{ 2 }-8x-2y-51=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+8x+2y-51=0$$

D
None of these

Solution

End points of diameter are $$(4,3)$$ and $$(-12,-1)$$
$$\therefore $$ Center of circle$$=\left( \cfrac { 4-12 }{ 2 } ,\cfrac { 3-1 }{ 2 } \right) $$
$$=(-4,1)$$
Radius$$=\cfrac { 1 }{ 2 } \sqrt { { \left( 4+12 \right) }^{ 2 }+{ \left( 3+1 \right) }^{ 2 } } $$
$$=\sqrt { 68 } $$
$$\therefore $$ Equation of circle is $${ \left( x+4 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( \sqrt { 68 } \right) }^{ 2 }$$
$${ x }^{ 2 }+8x+16+{ y }^{ 2 }-2y+1=68$$
$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$
Question 10
1 / -0

Axis of a parabola is $$y=x$$ and vertex and focus are at a distance $$\sqrt{2}$$ and $$2\sqrt{2}$$ respectively from the origin. The equation of the parabola is

A
$${ \left( x-y \right) }^{ 2 }=8\left( x+y-2 \right) $$

B
$${ \left( x+y \right) }^{ 2 }=2\left( x+y-2 \right) $$

C
$${ \left( x-y \right) }^{ 2 }=4\left( x+y-2 \right) $$

D
$${ \left( x+y \right) }^{ 2 }=2\left( x-y+2 \right) $$

Solution

Given that
Axis of parabola is $$x=y$$

$$\Rightarrow $$Vertex is at distance $$\sqrt 2$$ from origin and lies on the axis

Therefore, $$\Rightarrow vertex=(1,1)$$

$$\Rightarrow $$ Focus is at a distance $$2\sqrt 2$$ from origin and lies on the axis.

Therefore, $$\Rightarrow vertex=(2,2)$$

So, we can say that equation of $$Directrix\rightarrow x+y=0$$

Hence equation of parabola is

$$\Rightarrow (x-2)^2+(y-2)^2=(\dfrac{x+y}{\sqrt 2})^2$$

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

$$\Rightarrow 2x^2-8x+8+2y^2-8y+8=x^2+y^2+2xy$$

$$\Rightarrow x^2+y^2-2xy=8x+8y-16$$

$$\Rightarrow (x-y)^2=8(x+y-2)$$

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 33

Result

Conic Sections Test - 33

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SHARING IS CARING

Question 1

$$\dfrac43$$

$$\dfrac4{\surd 3}$$

$$\dfrac2{\surd 3}$$

$$None\ of\ these$$

Solution

Question 2

first quadrant

second quadrant

third quadrant

none of these

Solution

Question 3

$$\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 5/4 } =1$$

$$\dfrac { { x }^{ 2 } }{ 5/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$$

$$\dfrac { { x }^{ 2 } }{ 3/4 } +\dfrac { { y }^{ 2 } }{ 1/4 } =1$$

$$\dfrac { { x }^{ 2 } }{ 1/4 } +\dfrac { { y }^{ 2 } }{ 3/4 } =1$$

Solution

Question 4

$${ x }^{ 2 }+{ y }^{ 2 }+ax=0$$

$${ x }^{ 2 }+{ y }^{ 2 }\pm 2ya=0$$

$${ x }^{ 2 }+{ y }^{ 2 }\pm 2xa=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+ya=0$$

Solution

Question 5

$$(a+b,c+d)$$

$$9-a-b,-c-d)$$

$$\left(\dfrac{a+b}{2},\dfrac{c+d}{2}\right)$$

none of these

Solution

Question 6

$$y^2 = 2x - 4$$

$$x^2 = 4y - 8$$

$$y^2 = 4x - 8$$

None

Solution

Question 7

$$x^{2}+y^{2}+13y=0$$

$$3x^{2}+3y^{2}+13x+3=0$$

$$6x^{2}+6y^{2}-13y=0$$

$$x^{2}+y^{2}+13+3=0$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$${ x }^{ 2 }+{ y }^{ 2 }-8x-2y-51=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+8x-2y-51=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+8x+2y-51=0$$

None of these

Solution

Question 10

$${ \left( x-y \right) }^{ 2 }=8\left( x+y-2 \right) $$

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

$${ \left( x-y \right) }^{ 2 }=4\left( x+y-2 \right) $$

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

Solution

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