Conic Sections Test

Result

Conic Sections Test - 43

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equation of the circle passing through the points of intersection of the lines $$2x+y=0,\ x+y+3=0$$ and $$x-2y=0$$ is

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

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

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

D
$$x^{2}+y^{2}+x+7y=0$$

Solution
Question 2
1 / -0

Length of the latus rectum of the parabola $$25 [(x- 2)^{2}+(y-3)^{2}]=(3x-4y+7)^2 $$ is :

A
4

B
2

C
1/5

D
2/5

Solution

$$The\, given\, equation\, is\, \\ 25\left[ { { { \left( { x-2 } \right) }^{ 2 } }+{ { \left( { y-3 } \right) }^{ 2 } } } \right] ={ \left( { 3x-4y+7 } \right) ^{ 2 } } \\ Here, \\ focus\, of\, parabola\, \left( { 2,3 } \right) \\ Directrix \\ 3x-4y+7 $$

$$\\ Now,\, we\, know\, distan ce\, between\, directrix\, and\, \, focus\, is\, \, half\, of\, latus\, rectum. \\ so,\, distan ce\, between\, them\Rightarrow L=\left| { \dfrac { { 3\times 2-4\times 3+7 } }{ { \sqrt { { 3^{ 2 } }+{ 4^{ 2 } } } } } } \right| \\ L=\left| { \dfrac { { 6-12+7 } }{ { \sqrt { 25 } } } } \right| \\ \therefore L=\left| { \dfrac { 1 }{ 5 } } \right| \\ Then\, latus\, rectum=2L \\ =2\times \dfrac { 1 }{ 5 } \\ =\dfrac { 2 }{ 5 } $$

$$\\ Hence,\, the\, option\, D\, is\, the\, correct\, answer.$$
Question 3
1 / -0

The length of the latus rectum of the parabola $$x^{2}-4x-8y+12=0$$ is-

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

We have
$${x^2} - 4x - 8y + 12 = 0$$

$$ \Rightarrow {\left( {x - 2} \right)^2} + 8 - 8y = 0\,\,\,\,\,\left( {by\,taking\,common\,x} \right)$$
$$ \Rightarrow {\left( {x - 2} \right)^2} = 8\left( {y - 1} \right)$$

Comparing with $$X^2=L.R\times Y$$

So, Length of latus rectum $$=8$$
Hence, the option $$(C)$$ is the correct answer.
Question 4
1 / -0

The equation of the circle of a radius $$5$$ in the first quadrant which touches the x-axis and the line $$3x-4y=0$$ is

A
$$x^{2}+y^{2}-24x-y-25=0$$

B
$$x^{2}+y^{2}-20x-12y+144=0$$

C
$$x^{2}+y^{2}-16x-18y+64=0$$

D
$$x^{2}+y^{2}-30x-10y+255=0$$

Solution

Hence, Option (D) is the correct answer.
Question 5
1 / -0

Equation of circle touching $$x=0, y=0$$ and $$x=4$$ is

A
$$4(x^{2}+y^{2})-16x-16y+16=0$$

B
$$4(x^{2}+y^{2})-12x-12y+12=0$$

C
$$4(x^{2}+y^{2})-8x-8y+4=0$$

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

Solution

$$\textbf{Step -1: Finding the centre and radius from the given values.}$$
$$\text{We are given that}$$ $$(h,k)=(2,2),r=2$$
$$\textbf{Step -2: Finding the equation of the circle}$$
$$\text{Equation of a circle whose centre is}$$ $$(h,k)$$ $$\text{and radius is}$$ $$r$$ $$\text{is}$$
$$\Rightarrow (x-h)^2+(y-k)^2=r^2$$
$$\Rightarrow (x-2)^2+(y-2)^2=2^2$$
$$\Rightarrow x^2-4x+4+y^2+4y+4=4$$
$$\Rightarrow x^2+y^2-4x-4y=-4$$
$$\text{Multiplying throughout by}$$ $$4,$$
$$\Rightarrow 4(x^2+y^2)-16x-16y+16=0$$
$$\textbf{ Hence, correct option is A.}$$
Question 6
1 / -0

Find the equation of a circle whose center is (2,-1) and radius is 3

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

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

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

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

Solution

Equation of circle with center $$(h,k)$$ and radius $$r$$ is written as
$$(x-h)^2+(y-k)^2=r^2$$

then,
Equation of circle with center $$(2,-1)$$ and radius $$3$$ is written as
$$\Rightarrow (x-2)^2+(y+1)^2=3^2$$
$$\Rightarrow x^2+4-4x+y^2+1+2y=9$$
$$\therefore x^2+y^2-4x+2y-4=0$$
Question 7
1 / -0

The vertex and focus of a parabola are $$(-2,2),(-6,6)$$. Then its length of latus rectum is

A
$$8\sqrt {2}$$

B
$$4\sqrt {2}$$

C
$$16\sqrt {2}$$

D
$$12\sqrt {2}$$

Solution

Distance between focus and vertex
$$\begin{array}{l} focus\left( { -6,6 } \right) ,\, vertex\left( { -2,2 } \right) \\ a=\sqrt { { { \left( { -6+2 } \right) }^{ 2 } }{ { \left( { 6-2 } \right) }^{ 2 } } } \\ a=\sqrt { 16+16 } =4\sqrt { 2 } \end{array}$$
length latus vectum
$$\begin{array}{l} =4a=4\times 2\sqrt { 2 } \\ =16\sqrt { 2 } \end{array}$$
Question 8
1 / -0

The equation of the tangent to the curve y = 2sinx + sin2x at $$x=\frac { \pi }{ 3 } $$ on it is

A
$$(2,3)$$

B
$$y+\sqrt { 3 } =0$$

C
$$2t-3=0$$

D
$$2y-3\sqrt { 3 } =0$$

Solution

$${ (x-2) }^{ 2 }+{ (y-3) }^{ 2 }=0\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+9-6y=0\\ { x }^{ 2 }+{ y }^{ 2 }-4x+6y+13=0$$
radius$$=\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } $$
$$g=-2 \quad f=-3 \quad c=13$$
$$r=\sqrt { { (-2) }^{ 2 }+{ (-3) }^{ 2 }-13 } =\sqrt { 13-13 } =0$$
radius of the circle is zero .
Hence diameter$$=0$$
i.e., it is a point circle with centre at $$(2,3)$$.
Question 9
1 / -0

The order of the differential equation of the family of parabolas whose length of latus rectum is fixed and axis is the X-axis

A
$$2$$

B
$$1$$

C
$$3$$

D
$$4$$

Solution

$$\begin{array}{l} Equation\, can\, be\, { \left( { y-k } \right) ^{ 2 } }=a\left( { x-h } \right) \, \, \, a=fixed \\ differentiating\, this\, we\, get \\ 2\left( { y-k } \right) .y'=ax \\ \left( { y-k } \right) .{ y^{ n } }+y{ '^{ 2 } }=\frac { a }{ 2 } \\ \frac { { ax } }{ { 2y' } } .y''+y{ '^{ 2 } }=\frac { a }{ 2 } \\ ax.{ y^{ n } }+2y{ '^{ 3 } }=ay' \\ \therefore order\, =2 \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$$
Question 10
1 / -0

$${ \cos }^{ 4 }\dfrac { \pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 3\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 5\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 7\pi }{ 8 } =$$

A
$$\cfrac { 1 }{ 2 } $$

B
$$\cfrac { 3 }{ 2 } $$

C
$$\cfrac { 1 }{ 4 } $$

D
$$\cfrac { 3 }{ 4 } $$

Solution

$$\textbf{Step 1: Simplify the given expression.}$$

$$\text{Consider, } \cos^4\dfrac{\pi}{8} + \cos^4\dfrac{3\pi}{8} + \cos^4\dfrac{5\pi}{8} + \cos^4\dfrac{7\pi}{8}$$

$$= \cos^4\dfrac{\pi}{8} + \cos^4\left(\dfrac{\pi}{2}-\dfrac{\pi}{8}\right) + \cos^4 \left ( \dfrac{\pi}{2} + \dfrac{\pi}{8} \right ) + \cos^4 \left ( \pi - \dfrac{\pi}{8} \right )$$

$$= \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} + \left (-\sin\dfrac{\pi}{8} \right )^4 + \left ( -cos\dfrac{\pi}{8} \right )^4$$

$$= 2\left [ \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} \right]$$

$$\textbf{Step 2: Use multiple angle formula to get final result. }$$

$$= 2 \left \{ \left ( \cos^2\dfrac{\pi}{8} + \sin^2\dfrac{\pi}{8} \right )^2 - 2\cos^2\dfrac{\pi}{8} \cdot \sin^2\dfrac{\pi}{8} \right \}$$

$$= 2 \left ( 1^2 - \dfrac{1}{2} \left ( 2\sin\dfrac{\pi}{8}\cos\dfrac{\pi}{8} \right )^2 \right )$$

$$= 2\left [ 1^2 - \dfrac{1}{2} \times \left ( \sin\dfrac{\pi}{4} \right )^2 \right ]$$ $$[\because \boldsymbol{\sin 2A=2\sin A\cos A}]$$

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

$$= 2\left [1^2 - \dfrac{1}{4} \right ]=\dfrac{3}{2}$$

$$\textbf{Hence, Option 'B' is correct.}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Conic Sections Test - 43

Result

Conic Sections Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

4

2

1/5

2/5

Solution

Question 3

$$4$$

$$6$$

$$8$$

$$10$$

Solution

Question 4

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

$$x^{2}+y^{2}-20x-12y+144=0$$

$$x^{2}+y^{2}-16x-18y+64=0$$

$$x^{2}+y^{2}-30x-10y+255=0$$

Solution

Question 5

$$4(x^{2}+y^{2})-16x-16y+16=0$$

$$4(x^{2}+y^{2})-12x-12y+12=0$$

$$4(x^{2}+y^{2})-8x-8y+4=0$$

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

Solution

Question 6

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

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

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

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

Solution

Question 7

$$8\sqrt {2}$$

$$4\sqrt {2}$$

$$16\sqrt {2}$$

$$12\sqrt {2}$$

Solution

Question 8

$$(2,3)$$

$$y+\sqrt { 3 } =0$$

$$2t-3=0$$

$$2y-3\sqrt { 3 } =0$$

Solution

Question 9

$$2$$

$$1$$

$$3$$

$$4$$

Solution

Question 10

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 3 }{ 2 } $$

$$\cfrac { 1 }{ 4 } $$

$$\cfrac { 3 }{ 4 } $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.