Conic Sections Test

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Conic Sections Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The equation of parabola whose latus rectum is $$2$$ units, axis is $$x+y-2=0$$ and tangent at the vertex is $$x-y+4=0$$ is given by

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

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

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

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

Solution

$${\textbf{Step - 1: Find value of 'a' from Length of latus rectum,}}$$
$${\text{We know that length of latus rectum of any parabola is 4a.}}$$
$${\text{Given that 4a = 2 }} \Rightarrow {\text{a = }}\dfrac{1}{2}.$$
$${\text{Also given that the axis is x + y = 2}}.$$
$${\textbf{Step - 2: Use relation between Perpendicular distance from axis and Distance from tangent of a parabola.}}$$
$${\text{From parabola's formula we know that,}}$$
$$\Rightarrow {\left( {{\text{perpendicular distance from axis}}} \right)^2}{\text{ = 4a}}\left( {{\text{distance from tangent}}} \right).$$
$$ \Rightarrow {\text{ }}{\left( {\dfrac{{{\text{x + y}} - {\text{2}}}}{{\sqrt 2 }}} \right)^2} = 4\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{{\text{x}} - {\text{y + 4}}}}{{\sqrt 2 }}} \right).$$
$$ \Rightarrow {\left( {{\text{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\text{x}} - {\text{y + 4}}} \right).$$
$${\textbf{Hence, The equation is }}{\left( {{\textbf{x + y}} - 2} \right)^2} = 2\sqrt 2 \left( {{\textbf{x}} - {\textbf{y + 4}}} \right).(C)$$
Question 2
1 / -0

If major axis is the x-axis and passes through the points $$(4, 3)$$ and $$(6, 2)$$, then the equation for the ellipse whose centre is the origin is satisfies the given condition.

A
$$\displaystyle \frac{x^2}{52} + \frac{y^2}{13} =1$$

B
$$\displaystyle \frac{x^2}{13} + \frac{y^2}{52} =1$$

C
$$\displaystyle \frac{x^2}{52} - \frac{y^2}{13} =1$$

D
$$\displaystyle \frac{x^2}{13} - \frac{y^2}{52} =0$$

Solution

Major axis is x - axis
Let the equation of the ellipse be $$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} =1$$
$$(4,3)$$ and $$(6, 2)$$ lies on it
$$\displaystyle \frac{16}{a^2} + \frac{9}{b^2} =1$$ ........ $$(i)$$

$$\displaystyle \frac{36}{a^2} + \frac{4}{b^2} =1$$ ........ $$(ii)$$
Subtracting $$(ii)$$ from $$(i)$$, we get
$$\displaystyle \frac{-20}{a^2} + \frac{5}{b^2} =0$$
$$\Rightarrow 5a^2 = 20 b^2 \Rightarrow a^2 = 4b^2$$
Putting the value of $$a^2$$ in eqn. $$(i)$$
$$\displaystyle \frac{16}{4b^2} + \frac{9}{b^2} = 1$$
$$\therefore b^2 =13$$ and $$a^2 = 4b^2 = 4 \times 13 =52$$
$$\therefore $$ Equation of ellipse is $$\displaystyle \frac{x^2 }{52}+\frac{y^2}{13}=1$$
Question 3
1 / -0

$$(4-\mathrm{a})\mathrm{x}^{2}+(12-\mathrm{a})\mathrm{y}^{2}=\mathrm{a}^{2}-16\mathrm{a}+48$$ represents an ellipse. Then:

A
$$a<12$$

B
$$a<4$$

C
$$a>4$$ & $$a<12$$

D
$$a>0$$

Solution

LHS of equation,we can rewrite it as
$$\displaystyle { a }^{ 2 }-16a+48=(a-4)(a-12)=(4-a)(12-a)$$
Therefore the given equation can be written as
$$\displaystyle \frac { { x }^{ 2 } }{ 12-a } +\frac { { y }^{ 2 } }{ 4-a } =1$$
This will represent an ellipse ,If $$\displaystyle 4-a>0\Longrightarrow a<4$$
($$\displaystyle \ddot { . } 4-a$$ is the smaller of $$\displaystyle 12-a$$ )
So, taking intersection of $$\displaystyle 12-a$$ and $$\displaystyle 12-a>0.$$
Question 4
1 / -0

Latus rectum of a parabola is a ........ line segment with respect to the axis of the parabola through the focus whose endpoints lie on the parabola.

A
perpendicular

B
parallel

C
tilted

D
None of these

Solution

Consider the above image which shows that the latus rectum of a parabola is a line segment $$perpendicular $$ to the axis of the parabola, through the focus and whose end points lie on the parabola.
Question 5
1 / -0

If the equation of the incircle of an equilateral triangle is $${ x }^{ 2 }+{ y }^{ 2 }+4x-6y+4=0$$, then the equation of the circumcircle of the triangle is

A
$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-23=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y-23=0$$

D
None of these

Solution

Given equation of the incirle is $${ x }^{ 2 }+{ y }^{ 2 }+4x-6y+4=0$$
Its incenter is $$(-2,3)$$ and inradius $$=\sqrt { 4+9-4 } =3$$
Since in an equilateral triangle, the incenter and the circumcenter coincide,
$$\therefore$$ Circumcenter $$=(-2,3)$$
Also, in an equilateral triangle, circumradius $$=2($$ inradius$$)$$
$$\therefore$$ Circumradius $$=2.3=6$$
$$\therefore$$ The equation of the circumcircle is
$${ \left( x+2 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ \left( 6 \right) }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0$$
Question 6
1 / -0

The length of the latus rectum of the parabola
$$169\left \{ (x-1)^{2}+(y-3)^{2} \right \}=(5x-12y+17)^{2}$$ is

A
$$\dfrac{14}{11}$$

B
$$\dfrac{12}{13}$$

C
$$\dfrac{28}{13}$$

D
none of these

Solution

Given parabola may be written as, $$\displaystyle (x-1)^2+(y-3)^2 = \left(\frac{5x-12y+17}{\sqrt{5^2+12^2}}\right)^2$$

Thus focus of the parabola is $$(1,3)$$ and directrix is $$5x-12y+17$$

Now distance between directrix to the focus is $$ =\cfrac{5(1)-12(3)+17}{\sqrt{5^2+12^2}}=\cfrac{14}{13}$$

Hence length of latus rectum of the parabola is $$=2\times \cfrac{14}{13}=\cfrac{28}{13}$$
Question 7
1 / -0

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

A
a point

B
a pair of straight lines

C
a circle of non zero radius

D
none of these

Solution

$$x^{2}+y^{2}-2x+4y+5=0$$
$$(x-1)^{2}+(y+2)^{2} -5+5=0$$
$$\Rightarrow (x-1)^{2}+(y+2)^{2}=0$$
Since, radius is $$0$$, hence its a point

Alternative method:
Here, $$a=b=1$$
$$r=\sqrt{1+4-5}=0$$
Hence, a circle of radius $$0$$. So, its a point.
Question 8
1 / -0

The equation $$7y^2-9x^2+54x-28y-116=0$$ represents

A
a hyperbola

B
a parabola

C
an ellipse

D
a pair of straight lines

Solution

Given conic is $$7y^2-9x^2+54x-28y-116=0$$

Here $$a=-9, b=7, c=-116, f=-14, g=27, h=0$$

$$\therefore \Delta=abc+2fgh-af^2-bg^2-ch^2=-9.7.(-116)+0-9(14)^2-7(27)^2-0=441\neq 0$$

and $$h^2=0, ab=-63$$

clearly $$h^2> ab$$

we know that for a equation $$ax^2+by^2+2hxy2gx+2fy+c=0$$ it represents a hyperbola if $$Δ ≠ 0$$ and $$h² > ab$$

Hence given conic represent hyperbola.
Question 9
1 / -0

If the parabola $$y^{2}=4ax$$ passes through $$(3,\:2)$$ then the length of latus rectum is

A
$$\displaystyle \frac{1}{3}$$

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

C
$$1$$

D
$$\displaystyle \frac{4}{3}$$

Solution

If the parabola $$y^{2}=4ax$$ passes through $$(3,\:2)$$
therefore, $$4=4a(3)$$
$$\Rightarrow a=\dfrac{1}{3}$$
Therefore, length of latus rectum: $$l=4a=\dfrac{4}{3}$$
Question 10
1 / -0

The equation of the circle passing through the point $$(1, 1)$$ and having two diameters along the pair of lines $$x^{2}-y^{2}-2x+4y-3=0$$ is

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

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

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

D
$$none\ of\ these$$

Solution

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

$$\Rightarrow x^2-2x+1 = y^2-4y+4$$
$$\Rightarrow (x-1)^2 = (y-2)^2\Rightarrow x-1 = \pm (y-2)$$

Thus lines are $$x-y=-1$$ and $$x+y=3$$
Solving these lines we get the centre of the required circle which is $$(1, 2)$$

Thus radius of the circle is $$\sqrt{(1-1)^2+(1-2)^2} = 1$$

Hence required circle is $$(x-1)^2+(y-2)^2 = 1\Rightarrow x^2+y^2-2x-4y+4=0$$

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 17

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Conic Sections Test - 17

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Question 1

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

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

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

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

Solution

Question 2

$$\displaystyle \frac{x^2}{52} + \frac{y^2}{13} =1$$

$$\displaystyle \frac{x^2}{13} + \frac{y^2}{52} =1$$

$$\displaystyle \frac{x^2}{52} - \frac{y^2}{13} =1$$

$$\displaystyle \frac{x^2}{13} - \frac{y^2}{52} =0$$

Solution

Question 3

$$a<12$$

$$a<4$$

$$a>4$$ & $$a<12$$

$$a>0$$

Solution

Question 4

perpendicular

parallel

tilted

None of these

Solution

Question 5

$${ x }^{ 2 }+{ y }^{ 2 }+4x+6y-23=0$$

$${ x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y-23=0$$

None of these

Solution

Question 6

$$\dfrac{14}{11}$$

$$\dfrac{12}{13}$$

$$\dfrac{28}{13}$$

none of these

Solution

Question 7

a point

a pair of straight lines

a circle of non zero radius

none of these

Solution

Question 8

a hyperbola

a parabola

an ellipse

a pair of straight lines

Solution

Question 9

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{2}{3}$$

$$1$$

$$\displaystyle \frac{4}{3}$$

Solution

Question 10

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

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

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

$$none\ of\ these$$

Solution

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