Conic Sections Test

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Conic Sections Test - 53

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Weekly Quiz Competition

Question 1
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A Iight ray gets reflected

from the $$x = -2$$. If the reflected touches the circle $$x^2 + y^2 = 4$$ and point of incident is $$(-2,-4)$$, then equation of incident ray is

A
$$4y+3x+22=0$$

B
$$3y+4x+20=0$$

C
$$4y+2x+20=0$$

D
$$y+x+6=0$$
Question 2
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Find the equation of the circle with center at $$(-3,5)$$ and passes through the point $$(5,-1)$$

A
$${(x+3)}^{2}+{(y-5)}^{2}=100$$

B
$$(x-3)^2+(y-5)^2$$

C
$$(x+3)^2+(y-5)^2$$

D
None of the above

Solution

Here we know that $$(h,k)=(-3,5)$$ but we are not given the radius.
However, we can find the radius by using the distance formula.
Therefore,
$$r = \sqrt {(x_2 – x_1)^2 + (y_2 – y_1)^2}$$
$$r = \sqrt {(5-(-3) )^2 + (-1-5)^2}$$
$$r = \sqrt {(8 )^2 + (6)^2}$$
$$r = \sqrt {64+36}$$
$$r = \sqrt {100}$$
$$r=10$$

An equation of the circle with center $$(h,k)$$ and radius $$r$$ is
$$(x-h)^{ 2 }+(y-k)^{ 2 }=r^{ 2 }$$

Substitute the values in the equation of circle as follows:
$$(x-(-3))^{ 2 }+(y-5)^{ 2 }=10^{ 2 }$$
i.e., $$(x+3)^{ 2 }+(y-5)^{ 2 }=100$$

Therefore, the equation of the circle is $$(x+3)^{ 2 }+(y-5)^{ 2 }=100$$.
Question 3
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Normals $$AA,A{A}_{1}$$ and $$A{A}_{2}$$ are drawn to the parabola $${ y }^{ 2 }=8x$$ from the point $$A(h,0)$$. If triangle $$O{A}_{1}{A}_{2}$$ is equilateral , then the possible value of $$h$$ is

A
$$26$$

B
$$24$$

C
$$28$$

D
None of these
Question 4
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Consider a parabola $$P$$ which touches $$y = 0$$ at $$(1, 0)$$ and $$x = 0$$ at $$(0, 2)$$, then latus rectum of $$P$$ is,

A
$$\dfrac {9}{5\sqrt {5}}$$

B
$$\dfrac {16\sqrt {5}}{5}$$

C
$$\dfrac {16}{25}$$

D
$$\dfrac {8}{25}$$

Solution

Let $$(a,b)$$ be the focus and $$y=m{x}$$ be directrix

$$\implies \bigg(\dfrac{m{x}-y}{\sqrt{1+m^{2}}}\bigg)^{2}=(x-a)^{2}+(y-b)^{2}$$

Differentiating on both sides

$$\dfrac{(m{x}-y)(m-\dfrac{d{y}}{d{x}})}{1+m^{2}}=(x-a)+(y-b)\dfrac{d{y}}{d{x}}$$

$$x$$ axis is tangent at $$(1,0)$$

$$\dfrac{m(m)}{1+m^{2}}=1-a\implies a=\dfrac{1}{1+m^{2}}$$

$$y$$ axis is tangent at $$(0,2)$$

$$\dfrac{-2}{1+m^{2}}=b-2\implies b=\dfrac{2{m}^{2}}{1+m^{2}}$$

$$(1,0)$$ lies on parabola

$$\dfrac{(m)^{2}}{1+m^{2}}=(1-a)^{2}+b^{2}$$

Substituting $$a$$ and $$b$$ values

$$\implies (4{m^2}-1)(m^{2})=0\implies m=0,\pm \dfrac{1}{2}$$

For $$m=0$$ we get $$a=1,b=0$$ which means that the directrix cuts the parabola which is not possible so $$m=\pm \dfrac{1}{2}$$

$$\implies a=\dfrac{4}{5},b=\dfrac{2}{5}$$

So the focus is $$\bigg(\dfrac{4}{5},\dfrac{2}{5}\bigg)$$

the directrix is $$2{y}+x=0$$

The axis will be $$2{x}-y+c=0$$

Focus lies on the axis $$\implies c=-\dfrac{6}{5}$$

the point of intersection of axis and directrix is $$\bigg(\dfrac{12}{25},\dfrac{-6}{25}\bigg)$$

Vertex will be $$\bigg(\dfrac{16}{25},\dfrac{2}{25}\bigg)$$

Latus Rectum will be $$4$$ times the distance between focus and vertex which is equal to $$\dfrac{16\sqrt{5}}{5}$$

Hence option $$B$$ is the answer.
Question 5
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A circle touches the y-axis at $$(0, 2)$$ and has an intercept of $$4$$ units on the positive side of the x-axis. Then the equation of the circle is?

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

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

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

D
none of these

Solution

Let $$s: x^2 + y^2 + 2gx + 2fy +c = 0$$ be the equation of the circle

$$S(0,2) = 0$$

$$\implies 4 + 4f + c =0 -eq.1$$

Intercept on x-axis is given by

$$2\sqrt{g^2 - c} = 4$$

$$g^2 – c = 4 -eq.2$$

Since x = 0 is a tangent

Radius = perpendicular distance from the center to tangent

$$\sqrt {g^2 + f^2 - c} = \dfrac{|-g|}{\sqrt{1 + 0}$$

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

$$f^2 = c$$

Substituting in eq.1

$$4 + 4f + f^2 =0 \implies (f+2)^2 = 0 \implies f = -2 $$

$$\implies c = 4$$

From eq.2

$$g^2 – 4 = 4$$

$$g = \pm 2 \sqrt{2}$$

Equation of circle is

$$x^2 +y^2 \pm 4\sqrt 2 x – 4y + 4 = 0$$

$$\implies x^2 +y^2 – 4(\sqrt 2 x + y) + 4 = 0$$
Question 6
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Let $$S$$ be the focus of $${ y }^{ 2 }=4x$$ and a point $$P$$ be moving on the curve such that its abscissa is increasing at the rate of $$4 units/s$$. Then the rate of increase of the projection of $$SP$$ on $$x+y=1$$ when $$P$$ is at $$(4,4)$$ is

A
$$\sqrt { 2 } $$

B
$$-1$$

C
$$-\sqrt { 2 } $$

D
$$-3\sqrt { 2 } $$

Solution
Question 7
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O is the centre of a circle of diameter $$4$$cm and OABC is a square, if the shaded area is $$\displaystyle\frac{1}{3}$$ area of the square, then the side of the square is __________.

A
$$\pi\sqrt{3}$$cm

B
$$\sqrt{3\pi}$$cm

C
$$3\sqrt{\pi}$$cm

D
$$3\pi$$cm

Solution

$$O$$ is the center of the circle and the diameter is $$4$$ units.
Also area of the shaded region$$=\cfrac { 1 }{ 3 } $$ area of the square
Area of the shaded region $$=\pi { r }^{ 2 }\times \cfrac { 90 }{ 360 } \\ =\pi \left( 4 \right) \left( \cfrac { 1 }{ 4 } \right) \\ =\pi $$
Area of the square $$={ a }^{ 2 }=\pi \times 3=3\pi $$
$$\therefore a=\sqrt { 3\pi } $$cm
Question 8
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Directions For Questions

Consider the circle $$x^2 + y^2 = 9$$ and the parabola $$y^2 = 8x$$. They intersect at $$P$$ and $$Q$$ in the first and the fourth quadrants, respectively. Tangents to the circle at $$P$$ and $$Q$$ intersect the x-axis at $$R$$ and tangents to the parabola at $$P$$ and $$Q$$ intersect the x-axis at $$S$$.

...view full instructions

The ratio of the areas of the triangle $$PQS$$ and $$PQR$$ is :

A
$$1 : \sqrt2$$

B
$$1 : 2$$

C
$$1 : 4$$

D
$$1 : 8$$

Solution

Equation of parabola is $$y^2=8x$$

Equation of circle is $$x^2+y^2=9$$
To find point of intersection of both, put equation of parabola in equation of circle.

$$\therefore x^2+8x=9$$
$$\therefore x^{ 2 }+8x-9=0$$
$$\therefore x^{ 2 }+9x-x-9=0$$
$$\therefore x\left( x+9 \right) -1\left( x+9 \right) =0$$
$$\therefore \left( x+9 \right) \left( x-1 \right) =0$$
$$\therefore x=-9$$ and $$x=1$$

But, $$x\neq -9$$ as $$y^2$$ cannot be negative
$$\therefore x=1$$

From equation of parabola,
$${ y }^{ 2 }=8$$
$$\therefore { y }=\pm 2\sqrt { 2 }$$

Thus, points of intersection are $$P\left( 1,2\sqrt { 2 } \right)$$ and $$Q\left( 1,-2\sqrt { 2 } \right)$$

Consider tangent to the circle which intersects x axis at point R.
Equation of tangent is,
$$x{ x }_{ 1 }+y{ y }_{ 1 }=9$$
$$\therefore x\left( 1 \right) +y\left( 2\sqrt { 2 } \right) =9$$
$$\therefore x+2\sqrt { 2 } y=9$$

When $$y=0$$, $$x=9$$
Thus, coordinates of point R are $$R\left( 9,0 \right)$$

Consider tangent to the parabola which intersects x axis at point S.
Equation of tangent is,
$$y{ y }_{ 1 }=4a\left( x+{ x }_{ 1 } \right)$$
$$\therefore y\left( 2\sqrt { 2 } \right) =8\left( x+1 \right)$$

When $$y=0$$, $$x=-1$$
Thus, coordinates of point S are $$R\left( -1,0 \right)$$

Area of triangle PQS is,
$$A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times PQ\times SM$$
$$\therefore A\left( \triangle PQS \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 2$$
$$\therefore A\left( \triangle PQS \right) =4\sqrt { 2 }$$

Area of triangle PQR is,
$$A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times PQ\times RM$$
$$\therefore A\left( \triangle PQR \right) =\dfrac { 1 }{ 2 } \times 4\sqrt { 2 } \times 8$$
$$\therefore A\left( \triangle PQR \right) =16\sqrt { 2 }$$

$$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =\dfrac { 4\sqrt { 2 } }{ 16\sqrt { 2 } }$$

$$\therefore \dfrac { A\left( \triangle PQS \right) }{ A\left( \triangle PQR \right) } =1:4$$
Question 9
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Directions For Questions

Consider the ellipse $${E}_{1}:\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1,\left( a>b \right) $$. An ellipse $${E}_{2}$$ passes through the extremities of the major axis of $${E}_{1}$$ and has its foci at the ends of its minor axis.
Consider the following property:
Sum of focal distances of any point on an ellipse is equal to its major axis.

...view full instructions

If eccentricity of both ellipses are same, then their eccentricity is

A
$$2-\sqrt { 3 } $$

B
$$\sqrt { 2 } -1$$

C
$$\cfrac { 3-\sqrt { 5 } }{ 2 } $$

D
$$\cfrac { \sqrt { 5 } -1 }{ 2 } \quad $$
Question 10
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An endless inextensible string of length $$15$$m passes around the pins, A & B which are $$5$$m apart. This string is always kept tight and a small ring, R of negligible dimensions, inserted in this string is made to move in a path keeping all segments RA, AB, RB tight (as mentioned earlier). The ring traces a path, given by conic C, then.

A
Conic C is an ellipse with eccentricity $$\displaystyle\frac{1}{2}$$

B
Conic C is an hyperbola with eccentricity $$2$$

C
Conic C is an ellipse with eccentricity $$\displaystyle\frac{2}{3}$$

D
Conic C is a hyperbola with eccentricity $$\displaystyle\frac{3}{2}$$

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 53

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Conic Sections Test - 53

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SHARING IS CARING

Question 1

$$4y+3x+22=0$$

$$3y+4x+20=0$$

$$4y+2x+20=0$$

$$y+x+6=0$$

Question 2

$${(x+3)}^{2}+{(y-5)}^{2}=100$$

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

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

None of the above

Solution

Question 3

$$26$$

$$24$$

$$28$$

None of these

Question 4

$$\dfrac {9}{5\sqrt {5}}$$

$$\dfrac {16\sqrt {5}}{5}$$

$$\dfrac {16}{25}$$

$$\dfrac {8}{25}$$

Solution

Question 5

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

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

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

none of these

Solution

Question 6

$$\sqrt { 2 } $$

$$-1$$

$$-\sqrt { 2 } $$

$$-3\sqrt { 2 } $$

Solution

Question 7

$$\pi\sqrt{3}$$cm

$$\sqrt{3\pi}$$cm

$$3\sqrt{\pi}$$cm

$$3\pi$$cm

Solution

Question 8

$$1 : \sqrt2$$

$$1 : 2$$

$$1 : 4$$

$$1 : 8$$

Solution

Question 9

$$2-\sqrt { 3 } $$

$$\sqrt { 2 } -1$$

$$\cfrac { 3-\sqrt { 5 } }{ 2 } $$

$$\cfrac { \sqrt { 5 } -1 }{ 2 } \quad $$

Question 10

Conic C is an ellipse with eccentricity $$\displaystyle\frac{1}{2}$$

Conic C is an hyperbola with eccentricity $$2$$

Conic C is an ellipse with eccentricity $$\displaystyle\frac{2}{3}$$

Conic C is a hyperbola with eccentricity $$\displaystyle\frac{3}{2}$$

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