Conic Sections Test

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Conic Sections Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the circle having normal at $$(3, 3)$$ as $$y = x$$ and passing through $$(2, 2)$$ is:

A
$$x^2 + y^2 - 3x - 7y + 12 = 0$$

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

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

D
$$x^2 + y^2 - 5x - 5y + 12 = 0$$

Solution

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

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)

A
$$be$$

B
$$e$$

C
$$ab$$

D
$$ae$$

Solution

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$$=m=e$$
Question 3
1 / -0

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

A
$$1$$

B
$$\dfrac {3}{4}$$

C
$$0$$

D
$$-\dfrac {3}{4}$$

Solution

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
Question 4
1 / -0

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

A
$$1$$

B
$$\dfrac{3}{4}$$

C
$$0$$

D
$$-\dfrac{3}{4}$$

Solution

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

for a circle of $$a(x-\alpha )^{2}+6(y-\beta )^{2}=1$$ then $$(a=6)$$
hence
$$\dfrac{\lambda }{3}=\dfrac{1}{4}$$
$$\lambda =\dfrac{3}{4}$$
Question 5
1 / -0

The equation of the circle touches y axis and having radius $$2$$ units and centre is $$(-2, -3)$$?

A
$$x^2+y^2-4x-9y-4=0$$

B
$$x^2+y^2+4x+9y+4=0$$

C
$$x^2+y^2+4x+6y+9=0$$

D
$$x^2+y^2-4x-6y-9=0$$

Solution
Question 6
1 / -0

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:

A
$$5$$

B
$$7$$

C
$$9$$

D
$$4$$

Solution

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,
$$a^2 -c^2 = b^2$$
so
$$16 -c^2 = b^2...............(1) $$

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)
$$16-9=b^2\Rightarrow b^2=7$$
Question 7
1 / -0

The radius of the circle $$x^2+y^2-5x+2y+5=0$$ is

A
$$2$$

B
$$1$$

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

D
$$\dfrac{2}{3}$$

Solution

Given the equation of the circle is $$x^2+y^2-5x+2y+5=0$$ can be written as
or $$x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+y^2+2.y.1+1+5-1-\dfrac{25}{4}=0$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2+4-\dfrac{25}{4}=0$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\dfrac{9}{4}$$
or $$\left(x+\dfrac{5}{2}\right)^2+(y+1)^2=\left(\dfrac{3}{2}\right)^2$$.
From this equation it is clear that the centre of the circle is $$\left(-\dfrac{5}{2},-1\right)$$ and radius is $$\dfrac{3}{2}$$.
Question 8
1 / -0

Cantres of the three circles
$$x^{2}+y^{2}-4x-6y-14=0$$
$$x^{2}+y^{2}+2x+4y-5=0$$
and $$x^{2}+y^{2}-10x-16y+7=0$$

A
Are the vertices of a right triangle

B
The vertices of an isosceles triangle which is not regular

C
Vertices of a regular triangle

D
Are collinear

Solution

Centre of $$x^2+y^2-4x-6y-14=0\space is\space (-2,-3)$$
Centre of $$x^2+y^2+2x+4y-5=0\space is\space (1,2)$$
Centre of $$x^2+y^2-10x-16y+7=0\space is\space (-5,-8)$$

Let these centres be $$A(-2,-3);B(1,2);C=(-5,-8)$$

Now we find the lengths of $$AB, BC, CA$$

$$\Rightarrow AB=\sqrt{(-2-1)^2+(-3-2)^2}=\sqrt {34}$$

$$\Rightarrow BC=\sqrt{(-5-1)^2+(-8-2)^2}=\sqrt {136}$$

$$\Rightarrow CA=\sqrt{(-2+5)^2+(-3+8)^2}=\sqrt {34}$$

Here we can see that $$AB+CA=BC$$

Therefore $$ ABC $$ is a straight line.
Question 9
1 / -0

Coordinates of centre and radius of the circle $$(x-3)^2+(y+4)^2=25$$ are respectively

A
$$(3,4)$$ , $$25$$

B
$$(-3,4)$$ , $$5$$

C
$$(3,-4)$$ , $$5$$

D
$$(3,-4)$$ , $$25$$

Solution

$$(x-3)^{2}+(y+4)^{2}=25$$
$$(x-3)^{2}+(y-(-4))^{2}-(5)^{2}$$
$$(x-4)^{2}+(y-k)^{2}-(r)^{2}$$
$$r=5 \,\,(h,k)= (3,-4)$$
Question 10
1 / -0

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

A
$$x=q$$

B
$$x=p$$

C
$$2x=1$$

D
$$2x + 1=0$$

Solution

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.

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 32

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Conic Sections Test - 32

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Question 1

$$x^2 + y^2 - 3x - 7y + 12 = 0$$

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

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

$$x^2 + y^2 - 5x - 5y + 12 = 0$$

Solution

Question 2

$$be$$

$$e$$

$$ab$$

$$ae$$

Solution

Question 3

$$1$$

$$\dfrac {3}{4}$$

$$0$$

$$-\dfrac {3}{4}$$

Solution

Question 4

$$1$$

$$\dfrac{3}{4}$$

$$0$$

$$-\dfrac{3}{4}$$

Solution

Question 5

$$x^2+y^2-4x-9y-4=0$$

$$x^2+y^2+4x+9y+4=0$$

$$x^2+y^2+4x+6y+9=0$$

$$x^2+y^2-4x-6y-9=0$$

Solution

Question 6

$$5$$

$$7$$

$$9$$

$$4$$

Solution

Question 7

$$2$$

$$1$$

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

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

Solution

Question 8

Are the vertices of a right triangle

The vertices of an isosceles triangle which is not regular

Vertices of a regular triangle

Are collinear

Solution

Question 9

$$(3,4)$$ , $$25$$

$$(-3,4)$$ , $$5$$

$$(3,-4)$$ , $$5$$

$$(3,-4)$$ , $$25$$

Solution

Question 10

$$x=q$$

$$x=p$$

$$2x=1$$

$$2x + 1=0$$

Solution

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