MCQ Questions for Class 11 Engineering Maths Conic Sections Quiz 5

The equation of the circle whose two diameters are the lines

$x+y=4$ and

$x-y=2$ and which passes through

$(4 , 6 )$ is

0%
$x^2+y^2-6x-2y-16=0$
0%
$x^2+y^2-6x-2y=15$
0%
$x^2+y^2=0$
0%
$5(x^2+y^2)-4x=16$

Explanation

Two diameters are the lines,

$x+y=4$ ----- ( 1 )

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

So, first, we find the intersection point of diameters that is the center of the circle.

Adding equation ( 1 ) and ( 2 ) we get,

$\Rightarrow$

$2x=6$

$\Rightarrow$

$x=3$

Substituting value of

$x$ in equation ( 1 ),

$\Rightarrow$

$3+y=4$

$\Rightarrow$

$y=1$

So, the center of circle is

$(3,1).$

Now, circle is pass through

$(4,6)$

The equation of circle is

$(x-a)^2+(y-b)=r^2$ , where

$(a,b)$ co-ordinates of circle and

$r$ is radius.

Put center coordinates and pass through point

$\Rightarrow$

$(4-3)^2+(6-1)^2=r^2$

$\Rightarrow$

$1+25=r^2$

$\Rightarrow$

$r^2=26$

So, the equation of circle is,

$\Rightarrow$

$(x-3)^2+(y-1)^2=26$

$\Rightarrow$

$x^2-6x+9+y^2-2y+1=26$

$\Rightarrow$

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

If the eccentricities of the hyperbola

$\dfrac {x^{2}}{a^{2}} - \dfrac {y^{2}}{b^{2}} = 1$ and

$\dfrac {y^{2}}{b^{2}} - \dfrac {x^{2}}{a^{2}} = 1$ be

$e$ and

$e_{1}$ , then

$\dfrac {1}{e^{2}} + \dfrac {1}{e_{1}^{2}} =$ .

0%
$1$
0%
$2$
0%
$3$
0%
None of these

Explanation

Eccentricity of hyperbola is always > 1

Hence

$\dfrac{1}{e} < 1$ so

$\dfrac{1}{e}+\dfrac{1}{e_1} <2$

Eccentricity of

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}=1$ is

$e=\sqrt{1+\dfrac{b^2}{a^2}}$

In case of the second hyperbola the axes gets interchanged and eccentricity is defined as the deviation of the curve from being circular.

$\implies$

$e_1=\sqrt{1+\dfrac{a^2}{b^2}}$

$\implies$

$\dfrac{1}{e^2} + \dfrac{1}{e_1^2} =$

$\dfrac{a^2}{a^2+b^2} + \dfrac{b^2}{a^2+b^2}=1$

The equation of the circle having

$x-y-2=0$ and

$x-y+2=0$ as two tangents and

$x-y=0$ as diameter is

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

Explanation

Since, the equations of tangents

$x-y-2=0$ and

$x-y+2=0$ are parallel.
Therefore, distance between them

$=$ Diameter of the circle

$=\dfrac { \left| 2-\left( -2 \right) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } }$

$\left( \because \dfrac { { C }_{ 2 }-{ C }_{ 1 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right)$

$=\dfrac { 4 }{ \sqrt { 2 } } =2\sqrt { 2 }$
Hence, radius

$=\dfrac { 1 }{ 2 } \left( 2\sqrt { 2 } \right) =\sqrt { 2 }$
It is clear from the figure that centre lies on the origin.
Therefore, equation of circle is

${ \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left( \sqrt { 2 } \right) }^{ 2 }$ .

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=2$

The equation of the latus rectum of the parabola

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

0%
$2y + 3 = 0$
0%
$3y = 2$
0%
$2y = 3$
0%
$3y + 2 = 0$

Explanation

$x^2+4x+2y=0\rightarrow x^2+4x+4+2y-4=0\rightarrow (x+2)^2=2(2-y)$

Now, focus of parabola

$(x-h)^2=4p(y-k)$ is

$F(h,k+p)$

Here

$h=-2,p=-\dfrac{1}{2},k=2\rightarrow F(-2,\dfrac{3}{2})$

Equation of axis is

$x-h=0\rightarrow x+2=0$

Now latus rectum is perpendicular to axis and passes through

$F$ .

$\therefore$ Slope of latus rectum

$=0$

Equation will be

$y-\dfrac{3}{2}=0(x-(-2))\rightarrow y=\dfrac{3}{2}\rightarrow 2y=3$

Hence,

$(C)$ .

The curve represented by

$x = 3\cos t + 3\sin t$ and

$y = 4\cos t - 4\sin t$ is

0%
An ellipse
0%
A hyperbola
0%
A parabola
0%
None of these

Explanation

The equations of curve are

$x =3(\cos t + \sin t)$

$\Rightarrow \dfrac {x}{3} = \cos t + \sin t .... (i)$

and

$y = 4(\cos t - \sin t)$

$\Rightarrow \dfrac {y}{4}= (\cos t - \sin t) .... (ii)$

On squaring and adding (i) and (ii)

$\dfrac {x^{2}}{9} + \dfrac {y^{2}}{16} = 2 (\sin^{2}t + \cos^{2}t)$

$\Rightarrow \dfrac {x^{2}}{9} + \dfrac {y^{2}}{16} = 2$ Which is an ellipse
Hence choice (a) is correct answer.

Find the equation of the circle which touches the coordinate axes and whose centre lies on the line

$x - 2y = 3$ .

0%
$(x-1)^2+(y+1)^2=1$
0%
$(x+1)^2+(y+1)^2=1$
0%
$(x-1)^2+(y-1)^2=1$
0%
$(x+1)^2+(y-1)^2=1$

Explanation

Given:

The equation of line is $x - 2y = 3$

Substitute $x = 0$ in the given equation.

$0 - 2y = 3 \Rightarrow - 2y = 3 \Rightarrow y = - \dfrac{3}{2}$

Substitute $y = 0$ in the given equation.

$x - 2\left (0 \right ) = 3 \Rightarrow x = 3$

The points are $\left (3, 0 \right ) \, and \, \left (0, - \dfrac{3}{2} \right )$ .

Let the centre be $\left (a, -a \right )$ .

Substitute $\left (a, -a \right )$ in given equation.

$a + 2a = 3 \Rightarrow 3a = 3 \Rightarrow a = 1$

Centre $= \left (1, -1 \right )$

Radius $= 1$

The equation of a circle becomes $\left (x - 1 \right )^{2} + \left (y + 1 \right )^{2} = 1$ .

Hence, the required solution is found.

Consider a rigid square

$ABCD$ as in the figure with

$A$ and

$B$ on the

$x$ and

$y$ axis respectively. When

$A$ and

$B$ slide along their respective axes, the locus of

$C$ forms a part of.

0%
A circle
0%
A parabola
0%
A hyperbola
0%
An ellipse which is not a circle

Explanation

Let the coordinates of C be (h,k). So, the coordinates of A and B can be determined as shown. The locus of C will be an equation containing the terms h and k.

Now, length of AB is constant. Using the distance formula, we get:

$(0-h)^2 + ( \frac{k}{2} - 0)^2 = constant \implies h^2 + \frac{k^2}{4} = constant$

We see that this equation is similar to that of an ellipse, which is

$\frac{x^2}{a^2}+ \frac{y^2}{b^2} = constant$

So, we can conclude that the locus of C is an ellipse. Here,

$a=1$ and

$b=2$ .

Since

$a \neq b$ , it is not a circle.

Suppose the parabola

$(y - k)^{2} = 4(x - h)$ with vertex

$A$ , passes through

$O = (0, 0)$ and

$L = (0, 2)$ . Let

$D$ be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at

$P$ . Then

$\angle PDA$ is equal to

0%
$\tan^{-1} \dfrac {1}{19}$
0%
$\tan^{-1} \dfrac {2}{19}$
0%
$\tan^{-1} \dfrac {4}{19}$
0%
$\tan^{-1} \dfrac {8}{19}$

Explanation

Given Parabola:

$({ y-k) }^{ 2 }=4(x-h)\quad .............(i)$

Vertex

$A$ :

$(h,k)$

Focus

$S$ :

$(1+h,k)$

Passes through

$O(0,0)$

$=>({ 0+k })^{ 2 }=4(0-h)$

$=>{ k }^{ 2 }=-4h\quad ...............(ii)$

Also Passes through

$\angle (0,2)=>({ 2-k })^{ 2 }=4(0-h)\quad .................(iii)$

From

$(ii)$ and

$(iii)$ ,

${ k }^{ 2 }={ k }^{ 2 }+4-4k$

$=>k=1$ and

$h=\cfrac { -1 }{ 4 } ...................(iv)$

End point of

$\angle R:(1-\cfrac { 1 }{ 4 } ,1\pm 2),\quad D:(\cfrac { 3 }{ 4 } ,3)\quad and\quad (\cfrac { 3 }{ 4 } ,-1)$

Axis of parabola:

$y-k=0 =>y=1............(v)$

Point of intersection of

$y=1$ with

$y-axis$ ie.

$x=0$ is

$P:(0,1)$

Line Passing through

$PD$ :

$(y-1)=(\cfrac { 3-1 }{ \cfrac { 3 }{ 4 } } )(x-0)$

$:y=\cfrac { 8 }{ 3 } x+1$

Slope of

$PD$ ,

$m=\cfrac { 8 }{ 3 } ......................(vi)$

Line Passing through

$AD$ ,

$(y-1)=(\cfrac { 3-1 }{ \cfrac { 3 }{ 4 } +\cfrac { 1 }{ 4 } } )(x+\cfrac { 1 }{ 4 } )$

$=>y= 2x+\cfrac { 1 }{ 2 } +1$

Slope of

$AD= { m }_{ 2 }=2..........(vii)$

Angle between

$AD$ and

$PD=>\angle PDA={ tan }^{ -1 }\left| \cfrac { \cfrac { 8 }{ 3 } -2 }{ 1+\cfrac { 8 }{ 3 } (2) } \right|$

$\angle PDA={ tan }^{ -1 }\left| \cfrac { \cfrac { 2 }{ 3 } }{ \cfrac { 19 }{ 3 } } \right|$

$\angle PDA={ tan }^{ -1 }(\cfrac { 2 }{ 19 } )$ .

$e$ and

$e'$ be the eccentricities of a hyperbola and its conjugate, them

$\dfrac {1}{e^2} + \dfrac {1}{e'^{2}}$ is equal to :

0%
0
0%
1
0%
2
0%
None of these

Explanation

$e^2 = \dfrac {a^2 + b^2}{a^2}$ and

$e'^{2} = \dfrac {a^2 + b^2}{b^2}$

$\Rightarrow \dfrac {1}{e^2} + \dfrac {1}{e'^2} = \dfrac {a^2}{a^2+b^2} + \dfrac {b^2}{a^2 + b^2} = 1$

$Ay^{2} + By + Cx + D = 0, (ABC\neq 0)$ be the equation of a parabola, then

0%
Then length of the latus rectum is $\left |\dfrac {C}{A}\right |$
0%
The axis of the parabola is vertical
0%
The y-coordinate of the vertex is $-\dfrac {B}{2A}$
0%
The x-coordinate of the vertex is $\dfrac {D}{A} + \dfrac {B^{2}}{4AC}$

Explanation

When,

$x={ A }\prime { y }^{ 2 }+B\prime { y }^{ 2 }+C\prime \quad$ then length of

$LR$ is

$\cfrac { 1 }{ \left| A\prime \right| } \quad ..................(1)$

Given:

$A{ y }^{ 2 }+By+Cx+D=0$

Converting the given equation into already known from ie. equation

$(1)$ ,

$=>A{ y }^{ 2 }+By+D=-Cx$

$=>x=\cfrac { A }{ C } { y }^{ 2 }+\cfrac { B }{ -C } y+\cfrac { D }{ -C } \quad ...................(2)$

Comparing equation

$(2)$ with equation

$(1)$ ,

$A\prime =\cfrac { A }{ -C }$

Now length of

$LR=\cfrac { 1 }{ \left| A\prime \right| } =\cfrac { 1 }{ \left| \cfrac { A }{ -C } \right| } =\cfrac { C }{ A }$

The equation of the circle passing through

$(2,0)$ and

$(0,4)$ and having the minimum radius is

0%
$x^{2}+y^{2}=20$
0%
$x^{2}+y^{2}-2x-4y=0$
0%
$x^{2}+y^{2}=4$
0%
$x^{2}+y^{2}=16$

Explanation

Given points are

$(2,0)$ and

$(0,4)$

Therefore, equation of circle is

$(x-2)(x-0)+(y-0)(y-4)=0$

By expanding, we get

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

Option B is correct.

$A(5,-4)$ and

$B(7,6)$ are points in a plane, then the set of all points

$P(x,y)$ in the plane such that

$AP=PB=2:3$ is

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

The equation of the circle which passes through the points

$(2, 3)$ and

$(5)$ and the centre lies on the straight line

$y - 4x + 3 = 0$ , is

0%
$x^{2} + y^{2} + 4x - 10y + 25 = 0$
0%
$x^{2} - y^{2} - 4x - 10y + 25 = 0$
0%
$x^{2} + y^{2} - 4x - 10y + 16 = 0$
0%
$x^{2} + y^{2} - 14y + 8 = 0$

Explanation

Centre lies on the line

$y-4x+3=0$

Let

$x=h$

$\Rightarrow y=4h-3$

So the center is of the form

$(h,4h-3)$

Distance of centre from

$(2,3)$ and

$(4,5)$ will be equal

$\Rightarrow { (h-2) }^{ 2 }+{ (4h-3-3) }^{ 2 }={ (h-4) }^{ 2 }+{ (4h-3-5) }^{ 2 }\\ \Rightarrow h=2$

So the centre is

$(2,5)$

$r=\sqrt { { (2-2) }^{ 2 }+{ (5-3) }^{ 2 } } =2$

So the equation of the circle is

${ (x-2) }^{ 2 }+{ (y-5) }^{ 2 }={ 2 }^{ 2 }\\ { x }^{ 2 }+{ y }^{ 2 }-4x-10y+25=0$

The equation of the image of the circle

${ x }^{ 2 }+{ y }^{ 2 }+16x-24y+183=0$ by the line mirror

$4x+7y+13=0$ is:

0%
${ x }^{ 2 }+{ y }^{ 2 }+32x-4y+235=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+32x+4y-235=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+32x-4y-235=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$

Explanation

Equation of given circle:

$x^2+y^2++16x-24y+183=0$

$\textbf{Idea}$ ; Lets find the reflection of centre of this circle with respect to the given line equation. Then find the radius of given circle. After reflection also, the radius of circle does not change. Thus finally knowing the centre of reflected circle and its radius we can find equation of reflected circle.

First convert this equation to standard form of circle equation as follows:

$x^2+y^2++16x-24y+183=0$

$(x^2+16x+64)+(y^2-24y+144)-25=0$

$(x+8)^2+(y-12)^2-25=0$

$(x+8)^2+(y-12)^2=5^2$

Thus

$\text{Centre}= (-8,12)$

and

$\text{radius}=\sqrt{25}=5$

Let reflected centre have co-ordinates=

$(h,k)$

Now note that mid point of actual centre and reflected centre will definitely fall on line of reflection.

Mid-point of

$(h,k)$ and

$(-8,12)$ =

$\left (\dfrac{h-8}{2},\dfrac{k+12}{2}\right)$

Lets try to satisfy line equation with given point.

$4x+7y+13=0\rightarrow4\left (\dfrac{h-8}{2}\right)+7\left (\dfrac{k+12}{2}\right)+13=0$

$4h+7k+78=0$ .....(1)

Also slope of given line is perpendicular to slope of RC

Slope of given line

$=-4/7$

$m_1 m_2=-1$

$\dfrac{\dfrac{(k+12)}{2}-12} { \left (\dfrac{h-8}{2}\right) +8}\times\dfrac{-4}{7}=-1$

$\rightarrow7h-4k+104=0$ ...........(2)

Solving (1) and (2) we get,

$(h,k)=(-16,-2)$

Equation of circle with centre

$(-16,-2)$ and radius

$=5$ :

$(x+16)^2+(y+2)^2=5^2$

$x^2+y^2+32x+4y+235=0$

The equation of the circle, which is the mirror image of the circle,

${ x }^{ 2 }+{ y }^{ 2 }-2x=0$ , in the line,

$y=3-x$ is:

0%
${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+24=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-8x-6y+24=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad$
0%
${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12=0\quad$

Explanation

Centre of circle

$(1,0)$ , radius

$= 1$

Image of

$(1,0)$ w.r.t. Line

$x+y-3=0$ is

$\dfrac{x-1}{1} = \dfrac{y-0}{1} = \dfrac{-2(1+0-3)}{1+1}=2$

$\Rightarrow x=3, \,\,y=2$

Now circle with centre

$(3,2)$ and radius

$1$ is

$(x-3)^2+(y-2)^2=1$

$\Rightarrow x^2+9-6x+y^2+4-4y=1$

$\Rightarrow x^2+y^2-6x-4y+12=0$

Point

$\left( 0,\lambda \right)$ lies in the interior of circle

$x^2+y^2=c^2$ then

0%
$\lambda \in \left( -c,c\right)$
0%
$\lambda \in \left(0,c \right)$
0%
$\lambda \in \left(-c,0\right)$
0%
None.

Explanation

The given equation is

$x^2+y^2=c^2$

Condition of a point to be interior of circle

$\implies x^2+y^2<c^2\\\implies 0+\lambda ^2<c^2\\\lambda ^2<c^2\\\lambda\in(-c,c)$

If the lines

$3x - 4y - 7 = 0$ and

$2s - 3y - 5 = 0$ are two diameters of a circle of area

$49\pi$ square units, the equation of the circle is-

0%
$x^{2} + y^{2} + 2x - 2y - 62 = 0$
0%
$x^{2} + y^{2} - 2x + 2y - 62 = 0$
0%
$x^{2} + y^{2} - 2x + 2y - 47 = 0$
0%
$x^{2} + y^{2} + 2x - 2y - 47 = 0$

Explanation

Given: Area of circle=

$49\pi \ sq.units$

$\therefore$ Area of circle=

$\pi \ r^2= 49\pi \ sq.units$

where,

$r$ is the radius of the circle

$\pi \ r^2= 49\pi$

$r^2= 49$

$r= 7 \ units$

Given eqs of lines are,

$3x-4y-7=0$ and

$2x-3y-5=0$

Solving the equations, we get

$x=1 \ , \ y=-1$

The points of intersection of the lines is the center of the circle i.e.,

$(1,-1)$

We have the eqn for the circle with center

$(a,b)$ and radius

$r$ ,

$(x-a)^2+(y-b)^2=(r)^2$

$(x-1)^2+(y+1)^2=(7)^2$

$(x^2+1-2x)+(y^2+1+2y)=49$

$x^2+y^2+2-2x+2y=49$

$x^2+y^2-2x+2y=49-2$

$\therefore$ The equation of the given circle is,

$x^2+y^2-2x+2y-47=0$

Find the equation of the circle which passes through the point

$(1, 1)$ & which touches the

$x^2 + y^2 + 4x - 6y - 3 = 0$ at the point

$(2, 3)$ on it.

0%
$x^2 + y^2 + x - 6y + 3 = 0$
0%
$x^2 + y^2 + x - 6y - 3 = 0$
0%
$x^2 + y^2 + x + 6y + 3 = 0$
0%
$x^2 + y^2 + x - 3y + 3 = 0$

Explanation

Correction(in 3rd last step):

$x^2+y^2 ``+"x-6y=\dfrac{9}{4}-\dfrac{1}{4}+4-9$

Change

$-x$ to

$+x$ in following steps

1116776_877049_ans_3c694f28d94d4387840341166ff6a968.jpg

If the locus of the point

$(4t^2 - 1, 8t-2)$ represents a parabola then the equation of latus rectum is

0%
$x - 5 = 0$
0%
$2x - 7 = 0$
0%
$x + 5 = 0$
0%
$x -3 = 0$

Explanation

$x=4t^2-1,y=8t-2$

$x+1=4t^2,y+2=8t$

let

$X=x+1;Y=y+2$

then the new coordinates will be

$X=4t^2, Y=8t$

$Y^2=4aX$ is the general equation of a parabola

$(8t)^2=4a\times 4t^2$

$64t^2=16at^2\Rightarrow a=4$

Equation of latus rectum of

$Y^2=4aX$ is

$X=a$

But we know that

$X=x+1$ ,substituting that in

$X=a$

$x+1=4\Rightarrow x-3=0$

Find the locus of the point of intersection of the lines

$\sqrt{3}x-y-4\sqrt{3} \lambda=0$ and

$\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ for different values of

$\lambda$ .

0%
$4x^2-y^2=48$
0%
$x^2-4y^2=48$
0%
$3x^2-y^2=48$
0%
$y^2-3x^2=48$

Explanation

Let

$(h,k)$ be the point of intersection of the given lines. Then,

$\sqrt{3}h-k-4\sqrt{3} \lambda=0$ and

$\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$

$\sqrt{3}h-k=4\sqrt{3}\lambda$ and

$\lambda(\sqrt{3}h+k)=4\sqrt{3}$

$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$

$3h^2-k^2=48$

Hence, the locus of (h,k) is

$3x^2-y^2=48$ .

Find the equation of the circle which passes through the point (1, 1) & which touches the circle

$x^{2} + y^{2} + 4x - 6y - 3 = 0$ at the point

$(2, 3)$ on it.

0%
$x^{2} + y^{2} + x - 6y + 3 = 0$
0%
$x^{2} + y^{2} + x - 6y - 3 = 0$
0%
$x^{2} + y^{2} + x + 6y + 3 = 0$
0%
$x^{2} + y^{2} + 4x - 3y + 3 = 0$

Explanation

REF.Image.

Let's consider given circle

$S_{1} = 0$ &

equation of required circle

$as\, S_{2} =0$

$Eq^n$ of tangent at P(2,3) to

$S_{1} = 0$

$x = 2$

(because equation of CP is y = 3 and tangent at P is

$\perp ^{le}$ to

Required circle should touch

$S_{1} =0$ at

$P(2,3)$

$\Rightarrow$ tangent at

$P(2,3)$ to

$S_{1} = 0$ should be

common tangent for

$S_{1} = 0 \, \&\, S_{2} = 0$

$\Rightarrow$ Required circle should touch the line x = 2

at P(2,3)

$eq^n$ of

$S_{2} =0$ will belong to family of circles

$S+\lambda L = 0$ [ S is point circle of L is tangent at p]

i.e.

$(x-2)^2 + (y-3)^2 +\lambda (x-2) = 0$

It should satisfy (1,1)

$\Rightarrow (1-2)^2+(1-3)^2+ \lambda (1-2) = 0$

$\Rightarrow \lambda = 5$

equation of required circle

$(x-2)^2 + (y-3)^2 + 5(x-2) = 0$

$\Rightarrow x^2 +y^2 +x -6y +3 =0$

1173758_877105_ans_22ca5c1b184c4aafbacd057f55e37a35.jpg

The equation of the hyperbola whose foci are

$(6, 5), (-4, 5)$ and eccentricity

$5/4$ is?

0%
$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$
0%
$\displaystyle\frac{x^2}{16}-\frac{y^2}{9}=1$
0%
$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$
0%
$\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$

Explanation

Let the centre of hyperbola be

$(\alpha , \beta)$

$y=5$ line has the foci, it also has the major axis.

$\therefore \dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$

Midpoint of foci = centre of hyperbola

$\therefore \alpha =1, \beta =5$

Given,

$e=\dfrac{5}{4}$ .

We know that foci is given by

$(\alpha \pm ae, \beta)$

$\therefore \alpha +ae=6$

$\Rightarrow 1+\dfrac{5}{4}a=6\Rightarrow a=4$

Using

$b^2=a^2(e^2-1)$

$\Rightarrow b^2=16\left(\dfrac{25}{16}-1\right)=9$

$\boxed{\therefore\ Equation\ of\ hyperbola \Rightarrow\\ \dfrac{(x-1)^2}{16}-\dfrac{(y-5)^2}{9}=1}$

The equation of the parabola with vertex at (0, 0), axis along x-axis and passing through

$\displaystyle \left( \frac{5}{3}, \frac{10}{3} \right)$ is

0%
$y^2 = 20 x$
0%
$\displaystyle y^2 = \frac{20 x}{3}$
0%
$\displaystyle y^2 = \frac{10 x}{3}$
0%
$\displaystyle x^2 = \frac{20 y}{3}$

Explanation

The equation of parabola whose vertex is

$(0,0)$ and the axis is along x axis

$y^{2} = \pm 4ax$

as the curve is going through The point

$(5/3,10/3)$

So, the curve is

$y^{2} = 4ax$

$\therefore (10/3)^{2} = 4.a(5/3)$

$100 = 4.a.15$

$a = 5/3$

$\therefore$ The curve is

$y^{2} = 4.5/3.x$

$\therefore 3y^{2} = 20x$ [Ans]

1154934_874166_ans_5a987bed4c1b49bc86270eb7d1a7e9bb.jpg

$b$ and

$c$ are the lengths of the segments of any focal chord of a parabola

$y^{2} = 4ax$ , then length of the semi-latus rectum is

0%
$\dfrac{b+c}{2}$
0%
$\dfrac{bc}{b+c}$
0%
$\dfrac{2bc}{b+c}$
0%
$\sqrt{bc}$

Explanation

We know that, the semi latus-rectum of a parabola is the harmonic mean between the segments of any focal chord of the parabola.

Given that,

$b$ and

$c$ are lengths of the segments of any focal chord of the parabola

${y}^{2}=4ax$

So, Let

$l$ is the length of semi-latus rectum of the parabola.

Then,

$\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{l}$

$\Rightarrow\,l=\dfrac{2bc}{b+c}$

Hence, length of semi-latus-rectum

$=\dfrac{2bc}{b+c}$

If the curve y = | x- 3| touches the parabola

$y^2 = \lambda (x-4), \lambda >0$ , then latus rectum of the parabola, is

0%
2
0%
4
0%
8
0%
16

Explanation

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

The basic equation of parabola:

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

$\Rightarrow (h,k)=(4,0)$

As latus rectum of parabola is given by the equation: $|4a|$

$\Rightarrow 4a=\lambda$

Considering $a=1\Rightarrow \lambda =4$

$\therefore \lambda =4$

State whether following statements are true or false

Statement-1 : The only circle having radius

$\sqrt {10}$ and a diameter along line

$2x + y=5$ is

$x^2 + y^2 - 6x + 2y=0$ .
Statement-2: The line 2x + y=5 is a normal to the circle

$x^2+ y^2 - 6x + 2y =0.$

0%
Statement- 1 is false, statement-2 is true.
0%
Statement-1 is true,statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
0%
Statement-1 is true, statement-2 is false.
0%
Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Explanation

Statement 1 and 2 both are correct, but statement 2 is not correct explaination for statement 1.

$x^2+y^2-6x+2y=0$

$(-g,-f)=(-3,1), c=0$

$r^2=g^2+f^2-c$

$\therefore r^2=9+1-0=10$

$\Rightarrow r=\sqrt{10}$

$2x+y=5$ is also a normal to the circle.

If the curves

$f(x) = e^{x}$ and

$g(x) = kx^{2}$ touches each other then the value of

$k$ is equal to

0%
$2$
0%
$\dfrac {e^{2}}{4}$
0%
$\dfrac {e^{2}}{2}$
0%
$\dfrac {e}{2}$

The equation of the circle passing through

$(4,\ 5)$ having the centre at

$(2 ,\ 2)$ is

0%
$x^{2} + y^{2} + 4x + 4y-5=0$
0%
$x^{2} + y^{2} - 4x - 4y-5=0$
0%
$x^{2} + y^{2} -4x = 13$
0%
$x^{2} + y^{2} - 4x - 4y + 5=0$

Explanation

Let the equation of circle be

$x^{2}+y^{2}+2gx+2fy+c=0$

$\therefore$ as centre is

$(-g, -f)$

$g=-2, f=-2$ (

$\therefore$ as centre if given

$(2,2)$ )

$\therefore$ eqn is

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

Also circle passes through

$(4,5)$

$\therefore$ it satisfies

$(4,5)$

$\therefore (4)^{2}+(5)^{2}-4(4)-4(5)+C=0$

$\therefore 16+25-16-20+C=0$

$\therefore 5+C=0$

$\therefore C=-5$

Hence eqn of circle is

$x^{2}+y^{2}-4x-4y-5=0$

Let circles

${C}_{1}$ and

${C}_{2}$ an Argand plane be given by

$\left| z+1 \right| =3$ and

$\left| z-2 \right| =7\ \$ respectively. If a variable circle

$\left| z-{ z }_{ 0 } \right| =r\quad$ be inside circle

${C}_{2}$ such that it touches

${C}_{1}$ externally and

${C}_{2}$ internally then locus of

${z}_{0}$ describes a conic

$E$ whose eccentricity is equal to

0%
$\cfrac { 1 }{ 10 }$
0%
$\cfrac { 3 }{ 10 }$
0%
$\cfrac { 5 }{ 10 }$
0%
$\cfrac { 7 }{ 10 }$

Explanation

We have

${ C }_{ 1 }:{ \left( x+1 \right) }^{ 2 }+{ y }^{ 2 }=9$

${ C }_{ 2 }:{ \left( x-2 \right) }^{ 2 }+{ y }^{ 2 }=49$

Now

$C{ C }_{ 1 }=r+{ r }_{ 1 }$

and

$C{ C }_{ 2 }={ r }_{ 2 }-r$

$\Rightarrow C{ C }_{ 1 }+C{ C }_{ 2 }={ r }_{ 1 }+{ r }_{ 2 }$

Locus of C is an ellipse with focus at

${C}_{1}$ and

${C}_{2}$

Now

${ r }_{ 1 }+{ r }_{ 2 }=2a=10....(1)\quad$

and

${ d }_{ { C }_{ 1 }{ C }_{ 2 } }(focal\quad length)=2ae=3....(2)$

(1) and (2)

$\Rightarrow$ eccentricity

$e$ is

$\cfrac { 3 }{ 10 }$

The foci of an ellipse are located at the points

$(2, 4)$ and

$(2, -2)$ . The points

$(4, 2)$ lies on the ellipse. If

$a$ and

$b$ represent the lengths of the semi-major and semi-minor axes respectively, then the value of

$(ab)^{2}$ is equal to

0%
$68 + 22\sqrt {10}$
0%
$6 + 22\sqrt {10}$
0%
$26 + 10\sqrt {10}$
0%
$6 + 10\sqrt {10}$

Explanation

The distance between the foci is

$6$ , so

$c = 3$ .
The sum of the distance from

$(4, 2)$ to each of the foci is the major axis length,
so

$2a = \sqrt {(4 - 2)^{2} + (2 - 4)^{2}} + \sqrt {(4 - 2)^{2} + (2 + 2)^{2}}$

$= \sqrt {4 + 4} + \sqrt {4 + 16} = \sqrt {8} + \sqrt {20}$

$= 2\sqrt {2} + 2\sqrt {5} \Rightarrow a = \sqrt {2} + \sqrt {5}$
Also, for an ellipse,

$b^{2} = a^{2} - c^{2} = (\sqrt {2} + \sqrt {5})^{2} - 3^{2}$

$= 7 + 2\sqrt {10} = -2 + 2\sqrt {10}$ .
Thus, we have

$(ab)^{2} = (7 + 2\sqrt {10})(-2 + 2\sqrt {10})$

$= -14 + 14\sqrt {10} - 4\sqrt {10} + 40$

$= 26 + 10\sqrt {10}$ .

state whether following statements are true or false

Statement 1 :

$\sqrt {(x-1)^2 +y^2} + \sqrt{(x+1)^2 + y^2} = 4$ represent equation of ellipse
Statement 2 : The locus of point which moves such that sum of its distance from two fixed points is constant is an ellipse.

0%
Statement 1 is true, statement 2 is true and statement 2 is correct explanation for statement 1
0%
Statement 1 is true, statement 2 is true and statement 2 is NOT correct explanation for statement 1
0%
Statement 1 is true, statement 2 is false
0%
Statement 1 is false, statement 2 is true

Equation of the parabola having focus

$(3,2)$ and Vertex

$(-1,2)$ is

0%
$(x+1)^2=16(y-2)$
0%
$(x-1)^2=16(y+2)$
0%
$(y-2)^2=16(x+1)$
0%
$(y+2)^2=16(x-1)$

Explanation

Equation of a General Parabola having axis along the x-axis is given by

$y^2=4ax$

where

$a=$ distance between Vertex and Focus.

Here

$y$ coordinate in Focus and Vertex is the same.

Thus this parabola would be along direction of the x-axis.

Diatance between vertex and focus

$=\sqrt{(3-(-1))^2+(2-2)^2}=4$

thus equation of parabola is

$(y-2)^2=4.(4)(x-(-1))$

i.e.

$(y-2)^2=16(x+1)$

The equation of the circle whose radius is

$5$ units and which touches the circle,

${x}^{2}+{y}^{2}-2x-4y-20=0$ at the point

$\left(5,5\right)$ is

0%
${x}^{2}+{y}^{2}+18x+16y+120=0$
0%
${x}^{2}+{y}^{2}-18x-16y+120=0$
0%
${x}^{2}+{y}^{2}-18x-16y-120=0$
0%
${x}^{2}-{y}^{2}-18x-16y-120=0$

Explanation

The given circle is

${x}^{2}+{y}^{2}-2x-4y-20=0$ has its centre

$-2g=-2$ or

$g=1$

$-2f=-4$ or

$f=2$

$Thus, the centre is at$

$C\left(1,2\right)$

$and radius$

$=r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{1}^{2}+{2}^{2}-\left(-20\right)}=5$

$units We have$

${O}_{1}=\left(1,2\right)$

$and Let$

${O}_{2}=\left(\alpha,\beta\right)$

$be the centres$

$A\left(5,5\right)$

$is the mid-point of$

${O}_{1}{O}_{2}$

$\therefore \dfrac{1+\alpha}{2}=5, \dfrac{2+\beta}{2}=5$

$On simplifying ,we get$

$\alpha=9,\beta=8$

$The circle is$

${\left(x-9\right)}^{2}+{\left(y-8\right)}^{2}={5}^{2}$

$or$

${x}^{2}+{y}^{2}-18x-16y+120=0$ $

The equation

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

$a\in$

0%
$\left(2,\dfrac{3}{2}\right)\cup\left(\dfrac{3}{2},5\right)$
0%
$\left(2,\dfrac{3}{2}\right)$
0%
$\left(1,\dfrac{3}{2}\right)$
0%
$\left(\dfrac{3}{2},5\right)$

Explanation

Given

$\dfrac{x^2}{2-a}+\dfrac{y^2}{a-5}+1=0$

$\dfrac{x^2}{a-2}+\dfrac{y^2}{5-a}=1$

For the above equation to be an ellipse

$a-2\gt 0$ ,

$5-a\gt0$ and

$a-2 \neq 5-a$

$a\gt2$ ,

$a\lt5$ and

$a\neq \dfrac{7}{2}$

$a \space \epsilon \left(2,\dfrac{7}{2}\right) \bigcup \left(\dfrac{7}{2},5\right)$

The equation of the latus rectum of the parabola

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

0%
$2y+3=0$
0%
$3y=2$
0%
$2y=3$
0%
$3y+2=0$

Explanation

Equation of Latus rectum of

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

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

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

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

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

$=-4\times \cfrac { 1 }{ 2 } \left( y-2 \right)$

Latus rectum will be at a distance of

$a=\cfrac { 1 }{ 2 }$

Distance from x-axis

$=2-\cfrac { 1 }{ 2 } =1.5=\cfrac { 3 }{ 2 }$

So,

$y=\cfrac { 3 }{ 2 }$

943386_1015888_ans_49d96163b004469a8cfbee79ebedcd31.png

Find the equation of the circle which passes through the points

$(2,-2)$ and

$(3,4)$ . And whose centre lies on the line

$x+y=2$ .

0%
${ x }^{ 2 }+{ y }^{ 2 }+7x+3y+16=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+4x+5y-16=0$
0%
${ x }^{ 2 }+{ y }^{ 2 }+5x+8y+30=0$

Explanation

$(2,-2)$ passes through the circle.

${ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\\ 4+4+4g-4f+c=0\\ 4g+4f+8+c=0\longrightarrow (1)$
As

$(3,4)$ passes through the circle

$25+6g+8f+c=0\longrightarrow (2)$
As center of a circle

$(-g,-f)$

$-g-f=2\\ g+f=-2\longrightarrow (3)\\ 4g-4f+8+c=0\\ 4\left( g-f+2+\cfrac { c }{ 4 } \right) =0\\ g-f+2+\cfrac { c }{ 4 } =0\\ -4=\cfrac { c }{ 4 } \\ c=-16\\ 4g-4f-8=0\times 2\\ 8g-8f-16=0\\ 6g+8f+9=0\times 1\\ 6g+8f+9=0\\ 2g=-7\\ g=\cfrac { -7 }{ 2 } \\ f=\cfrac { 3 }{ 2 }$
Equation of circle

${ x }^{ 2 }+{ y }^{ 2 }+2\times \cfrac { 7 }{ 2 } x-2\times \cfrac { 3 }{ 2 } y-16=0$

${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$

$16{m}^{2}-8l-1=0,$ then equation of the circle having

$lx+my+1=0$ is a tangent is

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

Explanation

$\because 16{m}^{2}=8l+1$

$\Rightarrow 16\left({l}^{2}+{m}^{2}\right)=16{l}^{2}+8l+1={\left(4l+1\right)}^{2}$
or

$4\sqrt{\left({l}^{2}+{m}^{2}\right)}=\left|4l+1\right|$
or

$\dfrac{\left|4l+1\right|}{\sqrt{\left({l}^{2}+{m}^{2}\right)}}=4$

$\therefore$ Centre

$\equiv\left(4,0\right)$ and radius

$\equiv4$
Equation of circle is

${\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}={4}^{2}$

$\Rightarrow {x}^{2}+16-8x+{y}^{2}=16$
or

${x}^{2}-8x+{y}^{2}=0$ is the required equation of the circle

If the line

$y = x \sqrt{3} - 3$ cuts the parabola

$y^2 = x + 2$ at

$P$ and

$Q$ and if

$A$ be the points

$(\sqrt{3}, 0)$ , then

$AP.AQ$ is

0%
$\dfrac{2}{3} (\sqrt{3} + 2)$
0%
$\dfrac{4}{3} (\sqrt{3} + 2)$
0%
$\dfrac{4}{3} (2 -\sqrt{3})$
0%
$\dfrac{4}{6 - 3\sqrt{3}}$

Explanation

Parabola:

$y^2=x+2 ..........(1)$

Line:

$y=x\sqrt{3}-3 ...........(2)$

$A(\sqrt3 ,0)$ lies on

$(2)$ and at

$x-axis$ ,

$P$ with respect to

$A,$ is

$P=(\sqrt3+rcos60°, rsin60°)$

$=(\sqrt{3} +\cfrac{r} {2},\cfrac { \sqrt { 3r } }{ 2 } )$ (where

$r=PA)$

$P$ is on Parabola,

$=>({ \cfrac { \sqrt { 3 } r }{ 2 } })^{ 2 }=\sqrt { 3 } +\cfrac { r }{ 2 } +2$

$=>\cfrac { { 3r }^{ 2 } }{ 4 } =\sqrt { 3 } +2+\cfrac { r }{ 2 }$

$=> { 3 }r^{ 2 }-2r+4(-\sqrt { 3 } -2)=0$

Is quadratic equation, having two roots say

$r_1$ and

$r_2$ which are lengths of

$PA$ and

$AQ$

$AP.AQ=r_1.r_2$

$AP.AQ=\cfrac { 4 }{ 3 } (\sqrt { 3 } +2).$

1025152_1034099_ans_9b12c3a4766f455d9a48765abee07a82.png

The equation of the circle having normal at

$(3, 3)$ as

$y = x$ and passing through

$(2, 2)$ is:

0%
$x^2 + y^2 - 3x - 7y + 12 = 0$
0%
$x^2 + y^2 -4x - 6y + 12 = 0$
0%
$x^2 + y^2 - 6x - 4y + 12 = 0$
0%
$x^2 + y^2 - 5x - 5y + 12 = 0$

Explanation

${ x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ 2x+2y\cfrac { dy }{ dx } +2g+2f\cfrac { dy }{ dx } =0.\\ \cfrac { dy }{ dx } =\cfrac { -y-x }{ y+f. } \\ \cfrac { dy }{ dx } \quad at\quad (3,3).\\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { -y-3 }{ 3+f } \\ Slope\quad of\quad normal\quad at\quad this\quad point\quad =1\\ \Rightarrow -\cfrac { (g+3) }{ f+3 } (1)\quad =-1\\ \Rightarrow g+3=f+3.\\ \Rightarrow g=f.\\ \Rightarrow { x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ Circle\quad passes\quad through\quad (3,3)\& (2,2).\\ \Rightarrow g+g+bg+bg+c=0\\ \Rightarrow 12g=-18-c.\\ \Rightarrow 4+4+4g+4g+c=0.\\ 8g=-8-c.\\ -12g=-18-c\\ -4g=10\\ g=\cfrac { -5 }{ 2 } =f.\\ C=-8(g+1)\\ =-8(\cfrac { -5 }{ 2 } +1)\\ =-8\times \cfrac { -3 }{ 2 } \\ =+12\\ \therefore \quad Equation\quad :\\ { x }^{ 2 }+y^{ 2 }-5x-5y+12=0.\\ Ans.$

In the $xy$ plane, the segment with end points $(3,8)$ and $( -5,2)$ is the diameter of the circle. The point $(k,10)$ lies on the circle for:

0%
no value of $k$
0%
exactly one integral $k$
0%
exactly one non integral $k$
0%
two real values of $k$

Explanation

The centre of the given circle is,

$C\left( x,y \right)=\left( \dfrac{3+5}{2},\dfrac{8+2}{2} \right)=\left( 4,5 \right)$

The length of the diameter of the circle is,

$D=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( 2-8 \right)}^{2}}}=2\sqrt{10}\ units$

Therefore,

Radius $=\sqrt{10}\ units$

Therefore, the equation of the required circle is,

${{\left( x-4 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=10$

Since, the point $\left( k,10 \right)$ lies on the circle, we have

${{\left( k-4 \right)}^{2}}+{{\left( 10-5 \right)}^{2}}=10$

${{k}^{2}}-8k+16+25=10$

${{k}^{2}}-8k+31=0$

This equation has no real roots. Therefore, no such value of

$k$ exists.

If the equation

$\dfrac{\lambda (x+1)^2}{3}+\dfrac{(y+2)^2}{4}=1$ represents a circle then

$\lambda = ?$

0%
$1$
0%
$\dfrac{3}{4}$
0%
$0$
0%
$-\dfrac{3}{4}$

Explanation

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

for a circle of

$a(x-\alpha )^{2}+6(y-\beta )^{2}=1$ then

$(a=6)$

hence

$\dfrac{\lambda }{3}=\dfrac{1}{4}$

$\lambda =\dfrac{3}{4}$

A circle is dawn its centre on the line

$x+y=2$ to touch the line

$4x-3y+4=0$ and pass through the point

$(0,1)$ . Find the equation.

0%
$(x-1)^2+(y-1)^2=1$
0%
$(x-20)^2+(y+18)^2=841$
0%
$(x-21)^2+(y+19)^2=841$
0%
None

Explanation

Let the center of circle of circle be

$(a,b)$ and radius

$=r$

$\therefore x+y=2\Rightarrow a+b=2\,\,\,b=2-a$ ........(1)

Since circle touches the line:

$4x-3y+4=0$ , Hence

$\bot^r$

distance of center from line will be equal to the radius of circle, hence

$\dfrac{|4a-3b+4|}{\sqrt{4^2+(-3)^2}}=r\dfrac{|4a-3(2-a)|}{5}$

Again the distance of centre

$(a,b)$ from

$pt(0,1)$

will be equal to the radius of circle.

$\therefore \sqrt{(a-0)^2+(b-1)^2}=r\Rightarrow \sqrt{a^2+(2-a-1)^2}=r$

$a^2+1+a^2-2a=r^2\Rightarrow 2a^2-2a+1=r^2$

From (ii)

$2a^2-2a+1=\left(\dfrac{|7a-2|}{5}\right)^2$

$a^2-22a+21=0\,\,\Rightarrow a=1,21$

$a=1\Rightarrow b=1,r=1$

$a=21\Rightarrow b=-19,r=29$

$\therefore Eq^n$ of circle is

$2$

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

$(x-21)^2+(y+19)^2=29^2=841$

If the equation

$\dfrac { \lambda { \left( x+1 \right) }^{ 2 } }{ 3 } +\dfrac { { \left( y+2 \right) }^{ 2 } }{ 4 }=1$ represents a circle then

$\lambda=$

0%
$1$
0%
$\dfrac {3}{4}$
0%
$0$
0%
$-\dfrac {3}{4}$

Explanation

in a circle

$x^2,y^2$ coefficients must be same [it is the condition for the equation to be circle]

therefore

$\dfrac{\lambda}{3}=\dfrac{1}{4}$

$\Rightarrow \lambda=\dfrac 34$

option B

A circle of radius

$2$ lies in the first quadrant and touches both the axes of co-ordinates. Then the equation of the circle with centre

$(6, 5)$ and touching the above circle externally is

0%
$(x - 6)^2 + (y - 5)^2 = 4$
0%
$(x - 6)^2 + (y - 5)^2 = 9$
0%
$(x - 6)^2 + (y - 5)^2 = 36$
0%
none of these

Explanation

$(h,k)$ is the center and the radius is

$r$ then the equation of the circle is given by

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

Given that The center of the circle

$(h,k)=(6,5)$ and the radius

$r=2$

Therefore, the equation of the circle is

$(x-6)^2+(y-5)^2=4$

The foci of the ellipse

$\dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} =1$ and the hyperbola

$\dfrac{x^{2}}{144} - \dfrac{y^{2}}{81} =\dfrac{1}{25}$ coincide, then the value of

$b^{2}$ is:

0%
$5$
0%
$7$
0%
$9$
0%
$4$

Explanation

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,

$a^2 -c^2 = b^2$

$16 -c^2 = b^2...............(1)$

Equation of Hyperbola can also be written as

$\dfrac{x^2}{\dfrac{144}{25}}-\dfrac{y^2}{\dfrac{81}{25}}=1$

For the hyperbola, which must have its transverse axis on the x-axis, the equation

$c^2 - a^2 = b^2\Rightarrow c^2-\dfrac{144}{25}=\dfrac{81}{25}\Rightarrow c^2=\dfrac{225}{25}=9$

Putting this value in equation (1)

$16-9=b^2\Rightarrow b^2=7$

The radius of the circle

$x^2+y^2-5x+2y+5=0$ is

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

Explanation

Given the equation of the circle is

$x^2+y^2-5x+2y+5=0$ can be written as

$x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+y^2+2.y.1+1+5-1-\dfrac{25}{4}=0$

$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2+4-\dfrac{25}{4}=0$

$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\dfrac{9}{4}$

$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\left(\dfrac{3}{2}\right)^2$ .

From this equation it is clear that the centre of the circle is

$\left(-\dfrac{5}{2},-1\right)$ and radius is

$\dfrac{3}{2}$ .

The equation

$2x^2+3y^2-8x-18y+35=\lambda$ represents?

0%
A circle for all $\lambda$
0%
An ellipse if $\lambda < 0$
0%
The empty set if $\lambda > 0$
0%
A-point if $\lambda = 0$

Explanation

Given:

$2x^{2} + 3y - 8x - 18y + 35 = \lambda$

$2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda$

$2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = \lambda$

For $\lambda = 0$ , then

$2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = 0$

Thus, the point is $\left ( 2,3 \right )$ .

Hence, the correct option is ‘d’.

Coordinates of centre and radius of the circle

$(x-3)^2+(y+4)^2=25$ are respectively

0%
$(3,4)$ , $25$
0%
$(-3,4)$ , $5$
0%
$(3,-4)$ , $5$
0%
$(3,-4)$ , $25$

Explanation

$(x-3)^{2}+(y+4)^{2}=25$

$(x-3)^{2}+(y-(-4))^{2}-(5)^{2}$

$(x-4)^{2}+(y-k)^{2}-(r)^{2}$

$r=5 \,\,(h,k)= (3,-4)$

The parabola

$y = px^{2} + px + q$ is symmetrical about the line

0%
$x=q$
0%
$x=p$
0%
$2x=1$
0%
$2x + 1=0$

Explanation

The vertex of the parabola

$y = ax^2 + bx + c$ is

$\left(\dfrac{-b}{2a}, \dfrac{-D}{4a}\right)$ where

$D = b^2 - 4ac$

The vertex of the parabola

$y = px^2 + px + q$ is

$\left(\dfrac{-p}{2p}, \dfrac{4pq - p^2}{4p}\right) = \left(\dfrac{-1}{2}, \dfrac{4q - p}{4}\right)$

i.e., The parabola

$y = px^2 + px + q$ is symmetrical about the line

$x = \dfrac{-1}{2}$ or

$2x + 1 =0$

Optoin D is correct.

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is equal to (where

$e$ is the eccentricity of the hyperbola)

0%
$be$
0%
$e$
0%
$ab$
0%
$ae$

Explanation

Let a hyperbola ,

$\cfrac{x^{2}}{a^{2}}-\cfrac{y^{2}}{b^{2}}=1$ ---------1

Let the Eccentricity be e.

End point or extremities of LR are

$(ae,\cfrac{\pm b^{2}}{a})$

Tangent at L

$(ae, \cfrac{b^{2}}{a})$ will be

$\Rightarrow \cfrac{x(ae)}{a^{2}}-\cfrac{y(\cfrac{b^{2}}{a})}{b^{2}}=1$ (as tangent at

$(x_{1},y_{1})$ is

$\cfrac{xx_{1}}{a^{2}}-\cfrac{yy_{1}}{b^{2}}=1$ )

$\Rightarrow \cfrac{xe}{a}-\cfrac{y}{a}=1$

$\Rightarrow xe-y=a \Rightarrow y=xe-a$ --------2

General equation of a tangent in slope form

$y=mx+c$ ------3

Comparing Equation 2 & 3 , as they represent the tangents of Hyperbola,

$m=e$

The gradient i.e. slope of tangent

$=m=e$

1010529_1048576_ans_ade655962b79402cb7336faa6ca753a4.png

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Conic Sections Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.