Conic Sections Test

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Conic Sections Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

What is the equation of a circle with center (-3,1) and radius 7?

A
$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=7$$

B
$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=49$$

C
$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=7$$

D
$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=49$$

Solution

the general equation of a circle with center at (a,b) and radius r is
$$(x-a)^2+(y-b)^2=r^2$$
so substituting the values we get
the equation of the circle is $$(x+3)^2+(y-1)^2=7^2=49$$
Question 2
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An ambulance company provides services within an 80 mile radius of their headquarters If this service area is represented graphically with the headquarters located at the coordinates (0, 0) what is the equation that represents the service area?

A
$$\displaystyle x^{2}+y^{2}=80$$

B
$$\displaystyle \left ( x-0 \right )^{2}+\left ( y-0 \right )^{2}=80$$

C
$$\displaystyle x^{2}+y^{2}=1600$$

D
$$\displaystyle x^{2}+y^{2}=6400$$

Solution

the general equation of a circle with center at (a,b) and radius r is
$$(x-a)^2+(y-b)^2=r^2$$
so substituting the values we get the circle equation as
$$x^2+y^2=80^2=6400$$ option D
Question 3
1 / -0

Graph the circle
$${ \left( x-3 \right) }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=4$$

A

B

C

D

Solution

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

$$\implies center = (3,-4) , r = 2$$

Since $$C = (3,-4)$$
C lies in the fourth quadrant with radius $$2$$
Question 4
1 / -0

The point $$P(9/2, 6)$$ lies on the parabola $$y^2=4ax$$, then parameter of the point P is

A
$$\frac{3a}{2}$$

B
$$\frac{2}{3a}$$

C
$$\frac{2}{3}$$

D
$$\frac{3}{2}$$

Solution

As point $$P\left( \dfrac { 9 }{ 2 } ,6 \right) $$ lies on the parabola $${ y }^{ 2 }=4ax$$, it must satisfy the equation of parabola.

$$\therefore { 6 }^{ 2 }=4a\left( \dfrac { 9 }{ 2 } \right)$$

$$\therefore 36=18a$$

$$\therefore a=2$$

Now, point P which lies on parabola has coordinates $$P\left( a{ t }^{ 2 },2at \right)$$
Where, t is a parameter

$$\therefore a{ t }^{ 2 }=\dfrac { 9 }{ 2 }$$

$$\therefore 2{ t }^{ 2 }=\dfrac { 9 }{ 2 }$$

$$\therefore { t }^{ 2 }=\dfrac { 9 }{ 4 }$$

$$\therefore { t }=\dfrac { 3 }{ 2 }$$
Question 5
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Find the equation of the circle with center on x + y = 4 and 5x + 2y + 1 = 0 and having a radius of 3

A
$$\displaystyle x^{2}+y^{2}+6x-16y+64=0$$

B
$$\displaystyle x^{2}+y^{2}+8x-14y+25=0$$

C
$$\displaystyle x^{2}+y^{2}+6x-14y+49=0$$

D
$$\displaystyle x^{2}+y^{2}+6x-14y+36=0$$

E
none of these

Solution

Given,
Centre of circle is on $$x+y=4$$ and $$5x+2y+1=0$$
Thus,
$$x+y=4$$
$$=>y=4-x$$ (i)
$$5x+2y+1=0$$ (ii)
$$=>5x+2(4-x)+1=0$$
$$=>5x+8-2x+1=0$$
$$=>3x+9=0$$
$$=>x=-3$$
$$\therefore y=4-(-3)$$
$$=7$$
$$\therefore$$ Centre of circle is $$(-3,7)$$
Radius of circle=$$3$$ (given)
$$\therefore $$ Equation of circle is
$$(x-(-3))^2+(y-7)^2=3^2$$
$$=>(x+3)^2+(y-7)^2=9$$
$$=>(x^2+6x+9)+(y^2-14y+49)-9=0$$
$$=>x^2+6x+y^2-14y+49=0$$
Question 6
1 / -0

Find the equation of k for which the equation $$\displaystyle x^{2}+y^{2}+4x-2y-k=0$$ represents a point circle

A
$$5$$

B
$$-5$$

C
$$6$$

D
$$-6$$

E
none of these

Solution

Standard form of circle is
$$(x-x_1)^2+(y-y_1)^2=r^2$$
When $$(x_1,y_1)$$ is the centre of the circle and $$r$$ is the radius of the circle.
Now,given equation is,
$$x^2+y^2+4x-2y-k=0$$
$$=>x^2+4x+y^2-2y=k$$
$$=>x^2+2.x.2+2^2+y^2-2.y.1+1^1=k+2^2+1^1$$
$$=>(x+2)^2+(y-1)^2=k+4+1$$
$$=>(x+2)^2+(y-1)^2=k+5$$
Since it represents a point circle
$$\therefore k+5=0$$
$$=>k=-5$$
Question 7
1 / -0

The equation $$y^{2} + 4x + 4y + k = 0$$ represents a parabola whose latus rectum is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$y^{2} + 4x + 4y + k = 0$$
$$y^{2} + 2\times 2y + 4 - 4 + 4x + k = 0$$
$$(y + 2)^{2} = -4x - k + 4$$
$$(y + 2)^{2} = -4 \left (x - \dfrac {4 + k}{4}\right )$$
$$\therefore$$ Latus rectum $$= 4$$ units
Question 8
1 / -0

Equation of the parabola with its vertex at $$(1, 1)$$ and focus $$(3, 1)$$ is

A
$$(x 1)^2 = 8 (y - 1)$$

B
$$(y - 1)^2 = 8 (x - 3)$$

C
$$(y - 1)^2 = 8 (x - 1)$$

D
$$(x - 3)^2 = 8(y - 1)$$

Solution

As the vertex is$$(1,1)$$ and focus is $$(3,1)$$, whose ordinate is same its axis of symmetry is $$y = 1$$.
And as vertex is equidistant from foci and directrix, and latter is perpendicular to axis of symmetry.

Directrix is $$x =1$$

As parabola is the locus of a point whose distance from directrix $$x+1 = 0$$ and focus $$ (3,1)$$
Its equation is $$(x-3)^2+(y-1)^2= (x+1)^2$$
$$\Rightarrow x^2-6X+9+y^2-2Y+1 = x^2+2X+1$$
$$\Rightarrow y^2-2y+9 = 8x$$

$$\Rightarrow (y-1)^2 = 8(x-1)$$
So, option C is the answer.
Question 9
1 / -0

The equation of a circle which has a tangent $$3x+4y=6$$ and two normals given by $$(x-1)(y-2)=0$$ is

A
$$(x-3)^2+(y-4)^2=5^2$$

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

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

D
$$x^2+y^2-2x-4y+5=0$$

Solution

Equation of tangent $$=3x+4y=6$$ and two normals are $$(x-1)(y-2)=0$$
$$\Rightarrow x-1=0\,$$ and $$\, y-2=0$$
$$\therefore \text{Radius}=\displaystyle \frac{3(1)+4(2)-6}{\sqrt{9+16}}$$
$$=\displaystyle \frac{5}{5}=1$$
$$\therefore$$ Equation of the circle is
$$(x-1)^2+(y-2)^2=1\, $$ or$$ \, x^2+y^2-2x-4y+4=0$$
Question 10
1 / -0

The ends of the latus rectum of the conic $$x^{2} + 10x - 16y + 25 = 0$$ are

A
$$(3, -4), (13, 4)$$

B
$$(-3, -4), (13, -4)$$

C
$$(3, 4), (-13, 4)$$

D
$$(5, -8), (-5, 8)$$

Solution

Given that
$$x^{2} + 10x - 16y + 25 = 0$$
$$\Rightarrow (x + 5)^{2} = 16y$$
$$\Rightarrow X^{2} = 4AY$$,
where $$X = x + 5, A = 4, Y = y$$
The ends of the latus rectum are
$$(2A, A)$$ and $$(-2A, A)$$
$$\Rightarrow x + 5 = 2(4)$$
$$\Rightarrow x = -8 - 5 = 3, y = 4$$
and $$x + 5 = -2(4)$$
$$\Rightarrow x = -8 - 5 = -13, y = 4$$
$$\Rightarrow (3, 4)$$ and $$(-13, 4)$$.