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CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Maths
Conic Sections
Quiz 7
Equation of the circle of radius 5 whose centre lies on y-axis in first quadrant and passes through$$\left( {3,\,\,\,\,2} \right)$$ is
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$${x^2} + {y^2} - 12y + 11 = 0$$
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$${x^2} + {y^2} - 6y - 1 = 0$$
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$${x^2} + {y^2} - 8y + 3 = 0$$
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None of these
Explanation
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
$$x^2+y^2+36-12y=25\Rightarrow x^2+y^2-12y+11=0$$
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 :
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$$x^{2}+y^{2}-2x-4y-44=0$$
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$$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$$
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$$x^{2}+y^{2}-2x+4y-43=0$$
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$$none\ of\ these$$
Explanation
$$x^2+y^2-2x+4y-20=0$$
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 $$(7)^2$$ $$r=\dfrac{36}{5.14}=7$$
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
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$$x^{2}+y^{2}=9$$
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$$x^{2}+y^{2}=\dfrac {49}{4}$$
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$$x^{2}+y^{2}=\dfrac {81}{16}$$
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$$none\ of\ these$$
Explanation
$$x^2+y^2-5x+2y-5=0$$
$$\sqrt {\dfrac {25}{4}+1+25}$$
$$=\sqrt {\dfrac {49}{4}}$$
$$ (5/2, -1)$$
$$x^2+y^2=\left (\sqrt {\dfrac {49}{4}}\right)^2$$
$$x^2+y^2=\dfrac {49}{4}$$
vertices of an ellipse are $$(0,\pm 10)$$ and its eccentricity $$e=4/5$$ then its equation is
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$$90x^2-40y^2=3600$$
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$$80x^2+50y^2=4000$$
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$$36x^2+100y^2=3600$$
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$$100x^2+36y^2=3600$$
Explanation
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, $$\dfrac { { x }^{ 2 } }{ 36 } +\dfrac { { y }^{ 2 } }{ 100 } =1\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }=3600\\ \Rightarrow 100{ x }^{ 2 }+36{ y }^{ 2 }-3600=0$$
The name of the conic represent by the equation $$x^2+y^2-2y+20x+10=0$$ is
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a hyperbola
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an ellipse
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a parabola
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circle
Explanation
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
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
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$${x^2} + {y^2} - 6y - 7 = 0$$
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$${x^2} + {y^2} - 6y +7 = 0$$
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$${x^2} + {y^2} - 6y - 5 = 0$$
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$${x^2} + {y^2} - 6y + 5 = 0$$
Explanation
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 $$ = (\sqrt{7},0),(-\sqrt{7},0) $$
$$ R = \sqrt{7+9} = 4 $$
$$ (x-0)^{2}+(y-3)^{2} = 4^{2} $$
$$ x^{2}+y^{2}-6y+9 = 16 $$
$$ x^{2}+y^{2}-6y-7 = 0 $$
The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is $$e=2/3$$
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$$36x^2+20y^2=405$$
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$$20x^2+36y^2=405$$
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$$30x^2+22y^2=411$$
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$$22x^2+32y^2=409$$
Explanation
$$\begin{array}{l} e=\frac { 2 }{ 3 } =\sqrt { \frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ c } } \\ \Rightarrow { \left( { \frac { 2 }{ 3 } } \right) ^{ 2 } }=\frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ { { a^{ 2 } } } } \\ \Rightarrow \frac { { 4{ a^{ 2 } } } }{ a } ={ a^{ 2 } }-{ b^{ 2 } } \\ \Rightarrow { b^{ 2 } }={ a^{ 2 } }-4{ a^{ 2 } }=\frac { { 5{ a^{ 2 } } } }{ 9 } \to \left( i \right) \end{array}$$
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
$$ \Rightarrow \frac{{{x^2}}}{{\left( {\frac{{405}}{{20}}} \right)}} + \frac{{{y^2}}}{{\left( {\frac{{405}}{{36}}} \right)}} = - 1,20{x^2} + 36{y^2} = 405$$
The equation $$\dfrac { x ^ { 2 } } { 10 - a } + \dfrac { y ^ { 2 } } { 4 - a } = 1$$ represents an ellipse if
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$$a < 4$$
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$$a > 4$$
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$$4 < a < 10$$
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None of these
Explanation
$$\dfrac{x^2}{10-a}+\dfrac{y^2}{4-a}=1$$
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 $$a >\dfrac{83}{12}$$
Eccentricity of the hyperbola $$x^{2}-y^{2}=4$$ is
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$$2$$
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$$1$$
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$$\dfrac {3}{2}$$
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$$\sqrt {2}$$
If the latus rectum of an ellipse $$x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$$ $$1$$ is $$1 / 2 $$ then $$\varphi $$ is
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$$\pi / 2$$
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$$\pi / 6$$
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$$\pi / 3$$
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$$5$$ $$\pi/ 12$$
Explanation
Given $$x^2 tan^2 \phi + y^2 \, sec^2 \phi = 1$$
$$\rightarrow \dfrac{x^2}{(1/tan^2 \phi)} + \dfrac{y^2}{(1/sec^2 \phi)} = 1$$
$$a = \pm \dfrac{1}{tan \phi} , b = \pm \dfrac{1}{sec \phi}$$
and $$\rightarrow e^2 = 1 - \dfrac{b^2}{a^2}$$
$$\rightarrow e^2 = 1 - \dfrac{1/sec^2 \phi}{1/tan^2 \phi} = 1 - \dfrac{tan^2 \phi}{sec^2 \phi}$$
$$\rightarrow e^2 = 1 - sin^2 \phi = cos^2 \phi$$
length of latus rectum
$$(LL') = \dfrac{2 b^2}{a} = 2a (1 - e^2)$$
$$\rightarrow 2a (1 - cos^2 \phi) = 2a. sin^2 \phi = \dfrac{1}{2} $$ (Given)
$$\therefore 2. \dfrac{cos \phi}{sin \phi} sin^2 \phi = \dfrac{1}{2}$$
$$\rightarrow 2 cos \phi \, sin \phi = \dfrac{1}{2}$$
$$\rightarrow sin^2 \phi = \dfrac{1}{2} $$
$$\rightarrow 2 \phi = \dfrac{\pi}{6} , \dfrac{5 \pi}{6}$$
$$\therefore \phi = \dfrac{\pi}{12}$$ or
$$\phi = \dfrac{ 5 \pi}{12}$$
If a circle with centre $$(0,0)$$ touches the line $$5x+12y=1$$ then it equation will be
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$$169(x^{2}+y^{2})=1$$
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$$ (x^{2}+y^{2})=169$$
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$$16(x^{2}+y^{2})=1$$
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$$ (x^{2}+y^{2})=13$$
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 $$e^{2}_{1}+e^{2}_{2}=$$
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$$2$$
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$$3$$
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$$4$$
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$$1$$
A circle has its centre on the $$y-axis$$ and passes through the origin, touches another circle with centre $$(2,2)$$ and radius
2
, then the radius of the circle is
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$$1$$
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$$1/2$$
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$$1/3$$
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$$1/4$$
$$L L ^ { '}$$ is the latus rectum of an ellipse and $$\triangle S ^ { \prime } L L ^ { ' }$$ is an equilateral triangle. Then $$e =$$
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$$\dfrac { 1 } { \sqrt { 2 } }$$
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$$\dfrac { 1 } { \sqrt { 3 } }$$
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$$\dfrac { 1 } { \sqrt { 5 } }$$
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$$\dfrac { 1 } { \sqrt { 7 } }$$
Length of the latus rectum of the parabola $$\sqrt {x}+\sqrt {y}=\sqrt {a}$$ is
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$$a\sqrt {2}$$
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$$\dfrac {a}{\sqrt {2}}$$
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$$a$$
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$$2a$$
Explanation
Given that,
$$\sqrt {x} + \sqrt {y} = \sqrt {a}$$
$$\dfrac {\sqrt {x}}{\sqrt {a}} + \dfrac {\sqrt {y}}{\sqrt {a}} = 1$$
$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{a}} = 1 ..... (1)$$
But we know that
$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$$
is a parabola with focus
$$F = \dfrac {ab}{c^{3}}$$ where $$c$$
Directrix $$ax + by = 0$$ Since the semi lotus rectum of a parabola is the
distance from focus to directrix, we have
Latus rectum $$= \dfrac {2. a\left (\dfrac {ab^{2}}{c^{2}}\right ) + b\left (\dfrac {a^{2}b}{c^{2}}\right )}{\sqrt {a^{2} + b^{2}}}$$
$$= \dfrac {4a^{2}b^{2}}{c^{3}}$$
From equation (1) to
$$a = b$$ so,
Latus rectum $$= \dfrac {4a^{2} (a^{2})}{(\sqrt {a^{2} + a^{2}})^{3}}$$
$$= \dfrac {4a^{4}}{(\sqrt {2a^{2}})^{3}}$$
$$= \dfrac {4a^{4}}{(\sqrt {2}a)^{3}}$$
$$= \sqrt {2}a$$
Latus rectum $$= \sqrt {2}a$$
Hence, this is the answer.
If there is exactly one tangent at a distance of $$4$$ units from one of the focus of $$\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{a^{2}-16}=1,a > 4$$, the length of latus rectum is :-
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$$16$$
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$$\dfrac {8}{3}$$
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$$12$$
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$$15$$
The equation $$\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$$ represents an ellipse, if
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$$r>2$$
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$$r\in \left(2,\:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$$
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$$r>5$$
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$$r<2$$
Explanation
Equating the equation of the ellipse with the second-degree equation
$$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$$ with $$\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$$
we get $$A=\dfrac{1}{2-r}, B=0, C=\dfrac{1}{r-5},D=0,E=0$$ and $$F=1$$
For the second degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition $${B}^{2}-4AC<0$$ and also $$A\neq C$$
$$\Rightarrow {\left(0\right)}^{2}-4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$$
$$\Rightarrow -4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$$
$$\Rightarrow \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)>0$$
$$\Rightarrow \left(2-r\right)\left(r-5\right)<0$$
$$\Rightarrow \left(r-2\right)\left(r-5\right)>0$$
$$\Rightarrow r>2$$
A variable circle is drawn to touch the x-axis at the origin.The locus of the pole at the straight line $$6 x + m y + n = 0$$ w.r.t. the variable circle has the equation:-
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$$x ( m y - n ) - e y ^ { 2 } = 0$$
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$$x ( m y + n ) - e y ^ { 2 } = 0$$
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$$x ( m y - n ) + \ell y ^ { 2 } = 0$$
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none
General solution of the equation $$ y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$$ represents _____________.
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a straight line or hyperbola
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a straight line or parabola
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a parabola or hyperbola
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circles
Equation of circles which touch both the axes and whose centres are at a distance of $$2\sqrt {2}$$ units from origin are
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$$x^{2}+y^{2}\pm 4x\pm 4y+4=0$$
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$$x^{2}+y^{2}\pm 2x\pm 2y+4=0$$
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$$x^{2}+y^{2}\pm x\pm y+4=0$$
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$$x^{2}+y^{2}-4=0$$
Explanation
Since $$h=k$$ (radius )
$$\sqrt{h^{2}+k^{2}}=2\sqrt{2}$$
$$h^{2}+k^{2}=8$$
$$2h^{2}=8$$ (Taking
circle
$$1^{st}$$ Quad)
$$h=2$$
$$\therefore k=2$$ in
$$\therefore$$ center $$=(2,2)$$
Eqn. of circle having centre
$$(h,k)$$ & radius $$=r$$
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
$$\therefore (x-2)^{2}+ (y-2)^{2}=4$$
The equation of the circle in the first quadrature touching each coordinate axis at a distance of one unit from the origin
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$$x^{2}+y^{2}-2x-2y+1=0$$
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$$x^{2}+y^{2}-2x-1=0$$
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$$x^{2}+y^{2}-2x=0$$
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None of these
Explanation
We can clearly see from diagram that co-ordinates of centre of circle $$=(1,1)$$ & radius is unit
Thus equation of circle:-
$$(x-1)^{2}+(y-1)^{2}=1^{2}$$
$$x^{2}+y^{2}-2x-2y+1=0$$
The equation of the circle which circumscribes the triangle formed by the lines x + y + 3 = 0, x - y + 1 =0 and x = 3 is
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$$x^2+y^2-6x+2y-15=0$$
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$$3x^2+3y^2-9=0$$
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$$x^2+y^2+6x-2y+15=0$$
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None of these
The parametric form of the equation of the circle $${ x }^{ 2 }+{ y }^{ 2 }=9$$ is:
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$$x=\sqrt { 3 } \cos { \theta } ,$$ $$y=\sqrt { 3 } \sin { \theta } $$
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$$x=3\cos { \theta } ,\quad y=3\sin { \theta } $$
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$$x=-\sqrt { 3 } \cos { \theta } ,$$ $$y=-\sqrt { 3 } \sin { \theta } $$
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none of these
The length of the latus rectum of the parabola $${ 2y }^{ 2 }+3y+4x-2=0$$ is ________.
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$$\dfrac{ 3 }{ 2 } $$
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$$\dfrac{ 1 }{ 3 } $$
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$$2$$
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None of these.
The equation of a circle with centre at $$(1, -2)$$ and passing through the centre of the given circle $$x^2+y^2+2y-3=0$$, is?
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$$x^2+y^2-2x+4y+3=0$$
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$$x^2+y^2-2x+4y-3=0$$
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$$x^2+y^2+2x-4y-3=0$$
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$$x^2+y^2+2x-4y+3=0$$
The equation of circle whose centre is $$(1, -3)$$ and which touches the line $$2x-y-4=0$$, is
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$$5x^2+5y^2+10x+30y+49=0$$
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$$5x^2+5y^2+10x-30y-49=0$$
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$$5x^2+5y^2-10x+30y-49=0$$
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None of these
For the ellipse $$ {12x}^{2} +{4y}^{2} +24x-16y+25=0 $$
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centre is $$(-1,2) $$
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Length of axes are $$ {\sqrt {3}} and 1 $$
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eceentricity is $$ \sqrt {\cfrac {2} {3}} $$
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All of these
Explanation
Given,
$$12x^2+4y^2+24x-16y+25=0$$
$$\Rightarrow 12(x+1)^2+4(y-2)^2=3$$
$$\dfrac{(x+1)^2}{\frac{1}{4}}+\dfrac{(y-2)^2}{\frac{3}{4}}=1$$
$$\therefore a=\dfrac{1}{2},b=\dfrac{\sqrt 3}{2}$$
⇒ Centre $$ = (-1, 2)$$
Here $$b^2>a^2$$
⇒ eccentricity$$(e) = \sqrt {\dfrac {b^2-a^2}{b^2}} $$
$$= \sqrt {\dfrac {\dfrac 3 4 - \dfrac 1 4}{\dfrac 3 4}}=\sqrt{\dfrac 2 3}$$
Length of arcs,
length of major arc $$=2b=2\left ( \dfrac{\sqrt 3}{2} \right )=\sqrt 3$$
length of minor arc $$=2a=2\left ( \dfrac{1}{2} \right )=1$$
Option D is correct.
If a conic passing through origin has $$( 3,3 ) , ( - 4,4 )$$ as its focii, then
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auxillary circle is $$( 2 x + 1 ) ^ { 2 } + ( 2 y - 7 ) ^ { 2 } = 2$$
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auxillary circle is $$( 2 x + 1 ) ^ { 2 } + ( 2 y - 7 ) ^ { 2 } = 98$$
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auxillary circle is $$( 2 x + 1 ) ^ { 2 } + ( 2 y - 1 ) ^ { 2 } = 49$$
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auxillary circle is $$( 2 x + 1 ) ^ { 2 } + ( 2 y - 1 ) ^ { 2 } = 41$$
Equation of the ellipse whose minor axis is equal to the distance between foci and whose latus rectum is $$10 ,$$ is given by ____________.
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$$2 x ^ { 2 } + 3 y ^ { 2 } = 100$$
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$$2 x ^ { 2 } + 3 y ^ { 2 } = 80$$
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$$x ^ { 2 } + 2 y ^ { 2 } = 100$$
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none of these
The latus rectum of a parabola whose focal chord is $$PSQ$$ such that $$SP=3$$ and $$SQ=2$$, is given by
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$$\dfrac{24}{5}$$
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$$\dfrac{12}{5}$$
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$$\dfrac{6}{5}$$
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$$\dfrac{48}{5}$$
Explanation
Given, $$SP=3$$
$$SQ=2$$
By property of parabola
The harmonic mean of the lengths two parts into which the focus divides the focal chord is equal to length of semi-latus rectum.
Let length of lotus rectum be L
$$\dfrac{2SP\times SQ}{SP+SQ}=\dfrac{1}{2}$$
$$L=\dfrac{4\times SP\times SQ}{SP+SQ}=\dfrac{4\times 3\times 2}{3+2}=\dfrac{24}{5}$$.
The angle between the curves $$x^3-3xy^2=2$$ and $$3x^2y-y^3=2$$ is?
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{2}$$
The ratio of the ordinates of a point and its corresponding point is $$\frac { 2 \sqrt { 2 } } { 3 }$$ then eccentricity is ____________________.
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$$\frac { 1 } { 3 }$$
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$$\frac { 2 } { 3 }$$
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$$\frac { \sqrt { 2 } } { 3 }$$
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$$\frac { 2\sqrt { 2 } } { 3 }$$
The length of the latus rectum of the parabola $$4y^{2}+2x-20y+17=0$$ is:
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$$3$$
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$$6$$
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$$\dfrac{1}{2}$$
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$$9$$
The locus of the moving point $$P(x,y)$$ satisfying $$\sqrt { { \left( { x-1 } \right) }^{ 2 }+{ y }^{ 2 } } +\sqrt { { \left( { x+1 } \right) }^{ 2 }+({ y-{ \sqrt { 12 } ) }^{ 2 } } } =$$ a will be an ellipse if
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$$a<4$$
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$$a>2$$
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$$a>4$$
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$$a<2$$
A circle has radius $$3$$ units and its centre lies on the line $$y=x-1$$. Then the equation of this circle if it passes through the point $$(7, 3)$$, is?
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$$x^2+y^2-8x-6y=16$$
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$$x^2+y^2+8x+6y+16=0$$
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$$x^2+y^2-8x-6y-16=0$$
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None of these
$$ABCD$$ is a square with side $$a$$. If $$AB$$ and $$AD$$ are taken as positive coordinate axes then equation of circle circumscribing the square is
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$$x^{2}+y^{2}-ax-ay=0$$
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$$x^{2}+y^{2}+ax+ay=0$$
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$$x^{2}+y^{2}-ax+ay=0$$
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$$x^{2}+y^{2}+ax-ay=0$$
Explanation
Let the radius of circle be a/2, since it is
half of the diameter which is a. The centre of
the circle is going to occur at (a/2, a/2), since that
is also the centre of square
Using $$ (x-h)^{2}+(y-k)^{2} = r^{2}$$
where r is the radius and (h,k) is centre
$$ (x-a/2)^{2}+(y-a/2)^{2} = (a/2)^{2}$$
$$ x^{2}-\frac{2ax}{2}+\frac{a^{2}}{4}+y^{2}-\frac{2ay}{2}+\frac{a^{2}}{4} = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2}+\frac{a^{2}}{2}-ax - ay = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2} = ax+ay.$$
$$ x^{2}+y^{2} = a(x+y) \Rightarrow x^{2}+y^{2}-ax-ay = 0.$$
The latus rectum of the conic $${ 3x }^{ 2 }+{ 4y }^{ 2 }-6x+8y-5=0$$ is ________________________.
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$$3$$
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$$\dfrac { \sqrt { 3 } }{ 2 } $$
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$$\dfrac { 2 }{ \sqrt { 3 } } $$
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$$\dfrac { 4 }{ \sqrt { 3 } } $$
The circle passing through $$\left(t,1\right),\left(1,t\right)$$ and $$\left(t,t\right)$$ for all values of $$t$$ also passes through
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$$\left(0,0\right)$$
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$$\left(1,1\right)$$
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$$\left(1,-1\right)$$
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$$\left(-1,-1\right)$$
Explanation
The general equation of circle is
$$(x-h)^{2}+(y-k)^{2} = r^{2},$$ where h and k are the
coordinates
of center of circle.
if It passes through $$(1,t)$$ so $$ (1-h)^{2}+(t-k)^{2} = r^{2}$$ __ (1)
if It passes through $$(t,t)$$ so $$ (t-h)^{2}+(t-k)^{2} = r^{2}$$ __ (2)
if It passes through $$(t,1)$$ so $$ (t-h)^{2}+(1-k)^{2} = r^{2}$$ __ (3)
on solving eq (1) & (2)
$$ 1-h = t-h \Rightarrow t = 1 $$
It t = 1, then the circle passes through (1,1).
The area bounded by the parabola $${ y }^{ 2 }=4xy\quad $$ and its rectum is :-
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$$\frac { { 4 }^{ a } }{ 3 } sq-units$$
0%
$$\frac { { 8a }^{ 2 } }{ 3 } sq-units$$
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$$\frac { { 4a }\sqrt { a } }{ 3 } sq-units$$
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$$\frac { { 8a }\sqrt { a } }{ 3 } sq-units$$
Identify the types of cuves with represent by the equation $$\frac { { x }^{ 2 } }{ 1-r } -\frac { { y }^{ 2 } }{ 1+t } =1 $$, where $$r>1$$
is _______________.
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0%
An ellipse
0%
A hyperbola
0%
A circle
0%
None of these
The equation of the circle which passes through the point (3,-2) and (-2,0) and centre line 2x-y=3,is
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$${ x }^{ 2 }+{ y }^{ 2 }-3x-12y+2=0$$
0%
$${ x }^{ 2 }+{ y }^{ 2 }-3x+12y+2=0$$
0%
$${ x }^{ 2 }+{ y }^{ 2 }+3x+12y+2=0$$
0%
None of these
Explanation
Let $$x^2+y^2+2gx+2fy+c=0$$........(1) be the equation $$(3,-2)$$ lies on the circle.
Plugging in
$$13+6g-4f+c=0$$.........(2)
$$(-2,0)$$ lies on the circle lies on the circle
$$4-2g+c=0$$........(3)
centre $$(-g,-f)$$ lies on $$2x-y-3=0$$
$$-2g+f-3=0$$..........(4)
This (4) equation including the first and $$3$$ unknowns g,f, and c (Note $$x$$ and $$y$$ are considered constant) conditions for consistency given.
$$\begin{vmatrix} x^2+y^2& 2x& 2y& 1\\ 13& 6& -4& 1\\ 4& -4x& 0& 1\\ -3& -2& 1& 0\end{vmatrix}=0$$
After somehow operations and expanding (one can directly expand also)
$$\boxed{x^2+y^2+3x+12y+2=0}$$
The eq. of the circle which touches the exes of y at a distance of $$4$$ from the origin and cuts the intercepts of $$6$$ units from the axis of x is
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$$x^2 +y^2 \pm 10x \pm 8y + 16=0$$
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$$x^2 +y^2 \pm 5x \pm 4y + 16=0$$
0%
$$x^2 +y^2 \pm 10x \pm 8y - 16=0$$
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$$x^2 +y^2 \pm 5x \pm 4y - 16=0$$
Let $$P,\ Q,\ R,\ S$$ be the feet of perpendicular drawn from the point $$(1,\ 1)$$ the lines $$y=3x+4$$ and $$y=-3x+6$$ and their angle bisectors respectively, the equation of the circle whose extremities of a diameter are $$R$$ and $$S$$ is
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$$3x^{2}+3y^{2}+104x-110=0$$
0%
$$x^{2}+y^{2}+104x-110=0$$
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$$3x^{2}+3y^{2}-4x-18y+16=0$$
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$$x^{2}+y^{2}-4x-18+16=0$$
The length of the latus rectum of the parabola whose focus is (3,0) and directrix is 3x-4y-2=0 is
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12
0%
1
0%
4
0%
none of these
Explanation
focus $$\equiv (a,0)$$ . $$f\equiv (3,0)$$
$$\therefore a=3$$
Length of latus rectum $$= 4a$$
$$= 4\times 3$$
$$= 12$$ cm
A parabola passing through the point $$(-4,-2)$$ has its vertex at the origin and $$y-$$axis as its axis. The latus rectum of the parabola is
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$$6$$
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$$8$$
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$$10$$
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$$12$$
The equation of the latusrectum of the parabola $${ x }^{ 2 }+4x+2y=0$$ is:-
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3y-2=0
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3Y+2=0
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2Y-3=0
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2Y+3=0
The circle $$(x-3a)^{2}+y^{2}=8ax$$ intersects the parabola $$y^{2}=4ax$$ at the ends of the rectum of the parabola.
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True
0%
False
The equation of the circle having as a diameter, the chord $$x - y - 1 = 0$$ of the circle $$2x^2 + 2y^2 - 2x - 6y - 25 = 0$$, is
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$$x^2 + y^2 - 3x - y - \dfrac{29}{2} = 0$$
0%
$$2x^2 + 2y^2 + 2x - 5y - \dfrac{29}{2} = 0$$
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$$2x^2 + 2y^2 - 6x - 2y - 21 = 0$$
0%
None of these
Equation of the circle which is such that the length of the tangents to it from the points (1,0), (0, 2) and (3, 2) are $$1,\sqrt { 7 } $$ respectively is
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$$6({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$$
0%
$$9({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$$
0%
$$3({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$$
0%
$${ x }^{ 2 }+{ y }^{ 2 }-28x-5y+28=0$$
Explanation
The curve represented by $$Rs \left(\dfrac{1}{z}\right)=C$$ is (where $$C$$ is a constant and $$\neq 0$$)
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0%
Ellipse
0%
Parabola
0%
Circle
0%
Straight line
Explanation
$$\begin{array}{l} { { Re } }\, \, \left( { \frac { 1 }{ z } } \right) =c \\ { { Re } }\, \, \left( { \frac { 1 }{ { x+iy } } } \right) =c \\ { { Re } }\, \, \left( { \frac { { x-iy } }{ { { x^{ 2 } }+{ y^{ 2 } } } } } \right) =c \\ \frac { x }{ { { x^{ 2 } }+{ y^{ 2 } } } } =c \\ c\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) -{ x }=0 . \end{array}$$
Hence, this is represent circle.
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