Conic Sections Test

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Conic Sections Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The length of the latus rectum of the parabola $$4y^{2}+2x-20y+17=0$$ is:

A
$$3$$

B
$$6$$

C
$$\dfrac{1}{2}$$

D
$$9$$

Solution
Question 2
1 / -0

The locus of the moving point $$P(x,y)$$ satisfying $$\sqrt { { \left( { x-1 } \right) }^{ 2 }+{ y }^{ 2 } } +\sqrt { { \left( { x+1 } \right) }^{ 2 }+({ y-{ \sqrt { 12 } ) }^{ 2 } } } =$$ a will be an ellipse if

A
$$a<4$$

B
$$a>2$$

C
$$a>4$$

D
$$a<2$$

Solution
Question 3
1 / -0

A circle has radius $$3$$ units and its centre lies on the line $$y=x-1$$. Then the equation of this circle if it passes through the point $$(7, 3)$$, is?

A
$$x^2+y^2-8x-6y=16$$

B
$$x^2+y^2+8x+6y+16=0$$

C
$$x^2+y^2-8x-6y-16=0$$

D
None of these

Solution
Question 4
1 / -0

$$ABCD$$ is a square with side $$a$$. If $$AB$$ and $$AD$$ are taken as positive coordinate axes then equation of circle circumscribing the square is

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

B
$$x^{2}+y^{2}+ax+ay=0$$

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

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

Solution

Let the radius of circle be a/2, since it is
half of the diameter which is a. The centre of
the circle is going to occur at (a/2, a/2), since that
is also the centre of square
Using $$ (x-h)^{2}+(y-k)^{2} = r^{2}$$
where r is the radius and (h,k) is centre
$$ (x-a/2)^{2}+(y-a/2)^{2} = (a/2)^{2}$$
$$ x^{2}-\frac{2ax}{2}+\frac{a^{2}}{4}+y^{2}-\frac{2ay}{2}+\frac{a^{2}}{4} = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2}+\frac{a^{2}}{2}-ax - ay = \frac{a^{2}}{2}$$
$$ x^{2}+y^{2} = ax+ay.$$
$$ x^{2}+y^{2} = a(x+y) \Rightarrow x^{2}+y^{2}-ax-ay = 0.$$
Question 5
1 / -0

For the ellipse $$ {12x}^{2} +{4y}^{2} +24x-16y+25=0 $$

A
centre is $$(-1,2) $$

B
Length of axes are $$ {\sqrt {3}} and 1 $$

C
eceentricity is $$ \sqrt {\cfrac {2} {3}} $$

D
All of these

Solution

Given,

$$12x^2+4y^2+24x-16y+25=0$$

$$\Rightarrow 12(x+1)^2+4(y-2)^2=3$$

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

$$\therefore a=\dfrac{1}{2},b=\dfrac{\sqrt 3}{2}$$

⇒ Centre $$ = (-1, 2)$$

Here $$b^2>a^2$$

⇒ eccentricity$$(e) = \sqrt {\dfrac {b^2-a^2}{b^2}} $$

$$= \sqrt {\dfrac {\dfrac 3 4 - \dfrac 1 4}{\dfrac 3 4}}=\sqrt{\dfrac 2 3}$$

Length of arcs,

length of major arc $$=2b=2\left ( \dfrac{\sqrt 3}{2} \right )=\sqrt 3$$

length of minor arc $$=2a=2\left ( \dfrac{1}{2} \right )=1$$

Option D is correct.
Question 6
1 / -0

The latus rectum of the conic $${ 3x }^{ 2 }+{ 4y }^{ 2 }-6x+8y-5=0$$ is ________________________.

A
$$3$$

B
$$\dfrac { \sqrt { 3 } }{ 2 } $$

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

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

Solution
Question 7
1 / -0

The circle passing through $$\left(t,1\right),\left(1,t\right)$$ and $$\left(t,t\right)$$ for all values of $$t$$ also passes through

A
$$\left(0,0\right)$$

B
$$\left(1,1\right)$$

C
$$\left(1,-1\right)$$

D
$$\left(-1,-1\right)$$

Solution

The general equation of circle is
$$(x-h)^{2}+(y-k)^{2} = r^{2},$$ where h and k are the
coordinates of center of circle.
if It passes through $$(1,t)$$ so $$ (1-h)^{2}+(t-k)^{2} = r^{2}$$ __ (1)
if It passes through $$(t,t)$$ so $$ (t-h)^{2}+(t-k)^{2} = r^{2}$$ __ (2)
if It passes through $$(t,1)$$ so $$ (t-h)^{2}+(1-k)^{2} = r^{2}$$ __ (3)
on solving eq (1) & (2)
$$ 1-h = t-h \Rightarrow t = 1 $$
It t = 1, then the circle passes through (1,1).
Question 8
1 / -0

The ratio of the ordinates of a point and its corresponding point is $$\frac { 2 \sqrt { 2 } } { 3 }$$ then eccentricity is ____________________.

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

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

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

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

Solution
Question 9
1 / -0

Identify the types of cuves with represent by the equation $$\frac { { x }^{ 2 } }{ 1-r } -\frac { { y }^{ 2 } }{ 1+t } =1 $$, where $$r>1$$
is _______________.

A
An ellipse

B
A hyperbola

C
A circle

D
None of these

Solution
Question 10
1 / -0

The equation of the circle which passes through the point (3,-2) and (-2,0) and centre line 2x-y=3,is

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

B
$${ x }^{ 2 }+{ y }^{ 2 }-3x+12y+2=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }+3x+12y+2=0$$

D
None of these

Solution

Let $$x^2+y^2+2gx+2fy+c=0$$........(1) be the equation $$(3,-2)$$ lies on the circle.

Plugging in

$$13+6g-4f+c=0$$.........(2)

$$(-2,0)$$ lies on the circle lies on the circle

$$4-2g+c=0$$........(3)

centre $$(-g,-f)$$ lies on $$2x-y-3=0$$

$$-2g+f-3=0$$..........(4)

This (4) equation including the first and $$3$$ unknowns g,f, and c (Note $$x$$ and $$y$$ are considered constant) conditions for consistency given.

$$\begin{vmatrix} x^2+y^2& 2x& 2y& 1\\ 13& 6& -4& 1\\ 4& -4x& 0& 1\\ -3& -2& 1& 0\end{vmatrix}=0$$

After somehow operations and expanding (one can directly expand also)

$$\boxed{x^2+y^2+3x+12y+2=0}$$