MCQ Questions for 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

0%
${x^2} + {y^2} - 12y + 11 = 0$
0%
${x^2} + {y^2} - 6y - 1 = 0$
0%
${x^2} + {y^2} - 8y + 3 = 0$
0%
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 :

0%
$x^{2}+y^{2}-2x-4y-44=0$
0%
$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$
0%
$x^{2}+y^{2}-2x+4y-43=0$
0%
$none\ of\ these$

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

0%
$x^{2}+y^{2}=9$
0%
$x^{2}+y^{2}=\dfrac {49}{4}$
0%
$x^{2}+y^{2}=\dfrac {81}{16}$
0%
$none\ of\ these$

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

0%
$90x^2-40y^2=3600$
0%
$80x^2+50y^2=4000$
0%
$36x^2+100y^2=3600$
0%
$100x^2+36y^2=3600$

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

0%
a hyperbola
0%
an ellipse
0%
a parabola
0%
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

0%
${x^2} + {y^2} - 6y - 7 = 0$
0%
${x^2} + {y^2} - 6y +7 = 0$
0%
${x^2} + {y^2} - 6y - 5 = 0$
0%
${x^2} + {y^2} - 6y + 5 = 0$

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

$= (\sqrt{7},0),(-\sqrt{7},0)$

$R = \sqrt{7+9} = 4$

$(x-0)^{2}+(y-3)^{2} = 4^{2}$

$x^{2}+y^{2}-6y+9 = 16$

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

1165798_1201426_ans_640586a3ba71419cac983305b164259c.jpg

The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is

$e=2/3$

0%
$36x^2+20y^2=405$
0%
$20x^2+36y^2=405$
0%
$30x^2+22y^2=411$
0%
$22x^2+32y^2=409$

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

0%
$a < 4$
0%
$a > 4$
0%
$4 < a < 10$
0%
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

0%
$2$
0%
$1$
0%
$\dfrac {3}{2}$
0%
$\sqrt {2}$

If the latus rectum of an ellipse

$x ^ { 2 } \tan ^ { 2 } \varphi + y ^ { 2 } \sec ^ { 2 } \varphi =$

$1$ is

$1 / 2$ then

$\varphi$ is

0%
$\pi / 2$
0%
$\pi / 6$
0%
$\pi / 3$
0%
$5$ $\pi/ 12$

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

0%
$169(x^{2}+y^{2})=1$
0%
$(x^{2}+y^{2})=169$
0%
$16(x^{2}+y^{2})=1$
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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

$e^{2}_{1}+e^{2}_{2}=$

0%
$2$
0%
$3$
0%
$4$
0%
$1$

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

0%
$1$
0%
$1/2$
0%
$1/3$
0%
$1/4$

$L L ^ { '}$ is the latus rectum of an ellipse and

$\triangle S ^ { \prime } L L ^ { ' }$ is an equilateral triangle. Then

$e =$

0%
$\dfrac { 1 } { \sqrt { 2 } }$
0%
$\dfrac { 1 } { \sqrt { 3 } }$
0%
$\dfrac { 1 } { \sqrt { 5 } }$
0%
$\dfrac { 1 } { \sqrt { 7 } }$

Length of the latus rectum of the parabola

$\sqrt {x}+\sqrt {y}=\sqrt {a}$ is

0%
$a\sqrt {2}$
0%
$\dfrac {a}{\sqrt {2}}$
0%
$a$
0%
$2a$

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 :-

0%
$16$
0%
$\dfrac {8}{3}$
0%
$12$
0%
$15$

The equation

$\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$ represents an ellipse, if

0%
$r>2$
0%
$r\in \left(2,\:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$
0%
$r>5$
0%
$r<2$

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:-

0%
$x ( m y - n ) - e y ^ { 2 } = 0$
0%
$x ( m y + n ) - e y ^ { 2 } = 0$
0%
$x ( m y - n ) + \ell y ^ { 2 } = 0$
0%
none

General solution of the equation

$y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$ represents _____________.

0%
a straight line or hyperbola
0%
a straight line or parabola
0%
a parabola or hyperbola
0%
circles

Equation of circles which touch both the axes and whose centres are at a distance of

$2\sqrt {2}$ units from origin are

0%
$x^{2}+y^{2}\pm 4x\pm 4y+4=0$
0%
$x^{2}+y^{2}\pm 2x\pm 2y+4=0$
0%
$x^{2}+y^{2}\pm x\pm y+4=0$
0%
$x^{2}+y^{2}-4=0$

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

$=r$

$(x-h)^{2}+(y-k)^{2}=r^{2}$

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

1346560_1239899_ans_864b464e398647a3a23b07d13694914e.png

The equation of the circle in the first quadrature touching each coordinate axis at a distance of one unit from the origin

0%
$x^{2}+y^{2}-2x-2y+1=0$
0%
$x^{2}+y^{2}-2x-1=0$
0%
$x^{2}+y^{2}-2x=0$
0%
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:-

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

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

1345995_1239872_ans_e9e7fcff53464f328ba3fb7ae64a9035.png

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

0%
$x^2+y^2-6x+2y-15=0$
0%
$3x^2+3y^2-9=0$
0%
$x^2+y^2+6x-2y+15=0$
0%
None of these

The parametric form of the equation of the circle

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

0%
$x=\sqrt { 3 } \cos { \theta } ,$ $y=\sqrt { 3 } \sin { \theta }$
0%
$x=3\cos { \theta } ,\quad y=3\sin { \theta }$
0%
$x=-\sqrt { 3 } \cos { \theta } ,$ $y=-\sqrt { 3 } \sin { \theta }$
0%
none of these

The length of the latus rectum of the parabola

${ 2y }^{ 2 }+3y+4x-2=0$ is ________.

0%
$\dfrac{ 3 }{ 2 }$
0%
$\dfrac{ 1 }{ 3 }$
0%
$2$
0%
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?

0%
$x^2+y^2-2x+4y+3=0$
0%
$x^2+y^2-2x+4y-3=0$
0%
$x^2+y^2+2x-4y-3=0$
0%
$x^2+y^2+2x-4y+3=0$

The equation of circle whose centre is

$(1, -3)$ and which touches the line

$2x-y-4=0$ , is

0%
$5x^2+5y^2+10x+30y+49=0$
0%
$5x^2+5y^2+10x-30y-49=0$
0%
$5x^2+5y^2-10x+30y-49=0$
0%
None of these

For the ellipse

${12x}^{2} +{4y}^{2} +24x-16y+25=0$

0%
centre is $(-1,2)$
0%
Length of axes are ${\sqrt {3}} and 1$
0%
eceentricity is $\sqrt {\cfrac {2} {3}}$
0%
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

0%
auxillary circle is $( 2 x + 1 ) ^ { 2 } + ( 2 y - 7 ) ^ { 2 } = 2$
0%
auxillary circle is $( 2 x + 1 ) ^ { 2 } + ( 2 y - 7 ) ^ { 2 } = 98$
0%
auxillary circle is $( 2 x + 1 ) ^ { 2 } + ( 2 y - 1 ) ^ { 2 } = 49$
0%
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 ____________.

0%
$2 x ^ { 2 } + 3 y ^ { 2 } = 100$
0%
$2 x ^ { 2 } + 3 y ^ { 2 } = 80$
0%
$x ^ { 2 } + 2 y ^ { 2 } = 100$
0%
none of these

The latus rectum of a parabola whose focal chord is

$PSQ$ such that

$SP=3$ and

$SQ=2$ , is given by

0%
$\dfrac{24}{5}$
0%
$\dfrac{12}{5}$
0%
$\dfrac{6}{5}$
0%
$\dfrac{48}{5}$

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

$\dfrac{2SP\times SQ}{SP+SQ}=\dfrac{1}{2}$

$L=\dfrac{4\times SP\times SQ}{SP+SQ}=\dfrac{4\times 3\times 2}{3+2}=\dfrac{24}{5}$ .

1258876_1259894_ans_db69f189a90c480bbac2a4e1b785a95b.jpg

The angle between the curves

$x^3-3xy^2=2$ and

$3x^2y-y^3=2$ is?

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{2}$

The ratio of the ordinates of a point and its corresponding point is

$\frac { 2 \sqrt { 2 } } { 3 }$ then eccentricity is ____________________.

0%
$\frac { 1 } { 3 }$
0%
$\frac { 2 } { 3 }$
0%
$\frac { \sqrt { 2 } } { 3 }$
0%
$\frac { 2\sqrt { 2 } } { 3 }$

The length of the latus rectum of the parabola

$4y^{2}+2x-20y+17=0$ is:

0%
$3$
0%
$6$
0%
$\dfrac{1}{2}$
0%
$9$

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

0%
$a<4$
0%
$a>2$
0%
$a>4$
0%
$a<2$

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?

0%
$x^2+y^2-8x-6y=16$
0%
$x^2+y^2+8x+6y+16=0$
0%
$x^2+y^2-8x-6y-16=0$
0%
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

0%
$x^{2}+y^{2}-ax-ay=0$
0%
$x^{2}+y^{2}+ax+ay=0$
0%
$x^{2}+y^{2}-ax+ay=0$
0%
$x^{2}+y^{2}+ax-ay=0$

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

$(x-a/2)^{2}+(y-a/2)^{2} = (a/2)^{2}$

$x^{2}-\frac{2ax}{2}+\frac{a^{2}}{4}+y^{2}-\frac{2ay}{2}+\frac{a^{2}}{4} = \frac{a^{2}}{2}$

$x^{2}+y^{2}+\frac{a^{2}}{2}-ax - ay = \frac{a^{2}}{2}$

$x^{2}+y^{2} = ax+ay.$

$x^{2}+y^{2} = a(x+y) \Rightarrow x^{2}+y^{2}-ax-ay = 0.$

1200308_1271569_ans_c72734a671ca449a9f2da06fe94dc934.jpg

The latus rectum of the conic

${ 3x }^{ 2 }+{ 4y }^{ 2 }-6x+8y-5=0$ is ________________________.

0%
$3$
0%
$\dfrac { \sqrt { 3 } }{ 2 }$
0%
$\dfrac { 2 }{ \sqrt { 3 } }$
0%
$\dfrac { 4 }{ \sqrt { 3 } }$

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

0%
$\left(0,0\right)$
0%
$\left(1,1\right)$
0%
$\left(1,-1\right)$
0%
$\left(-1,-1\right)$

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 :-

0%
$\frac { { 4 }^{ a } }{ 3 } sq-units$
0%
$\frac { { 8a }^{ 2 } }{ 3 } sq-units$
0%
$\frac { { 4a }\sqrt { a } }{ 3 } sq-units$
0%
$\frac { { 8a }\sqrt { a } }{ 3 } sq-units$

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 _______________.

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

0%
${ x }^{ 2 }+{ y }^{ 2 }-3x-12y+2=0$
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

0%
$x^2 +y^2 \pm 10x \pm 8y + 16=0$
0%
$x^2 +y^2 \pm 5x \pm 4y + 16=0$
0%
$x^2 +y^2 \pm 10x \pm 8y - 16=0$
0%
$x^2 +y^2 \pm 5x \pm 4y - 16=0$

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

0%
$3x^{2}+3y^{2}+104x-110=0$
0%
$x^{2}+y^{2}+104x-110=0$
0%
$3x^{2}+3y^{2}-4x-18y+16=0$
0%
$x^{2}+y^{2}-4x-18+16=0$

The length of the latus rectum of the parabola whose focus is (3,0) and directrix is 3x-4y-2=0 is

0%
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

0%
$6$
0%
$8$
0%
$10$
0%
$12$

The equation of the latusrectum of the parabola

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

0%
3y-2=0
0%
3Y+2=0
0%
2Y-3=0
0%
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.

0%
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

0%
$x^2 + y^2 - 3x - y - \dfrac{29}{2} = 0$
0%
$2x^2 + 2y^2 + 2x - 5y - \dfrac{29}{2} = 0$
0%
$2x^2 + 2y^2 - 6x - 2y - 21 = 0$
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

0%
$6({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$
0%
$9({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$
0%
$3({ x }^{ 2 }+{ y }^{ 2 })-28x-5y+28=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-28x-5y+28=0$

The curve represented by

$Rs \left(\dfrac{1}{z}\right)=C$ is (where

$C$ is a constant and

$\neq 0$ )

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.

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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