Conic Sections Test

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Conic Sections Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The area of the triangle formed by the tangent and the normal to the parabola $${ y }^{ 2 }=4ax,$$ both drawn at the same end of the latus rectum and the axis of the parabola is

A
$$2\sqrt { 2 } { a }^{ 2 }$$

B
$$2{ a }^{ 2 }$$

C
$$4{ a }^{ 2 }$$

D
None of these

Solution
Question 2
1 / -0

The latus rectum of the hyperbola $$16{x^2} - 9{y^2} = 144$$ is-

A
$$\dfrac{13}{6}$$

B
$$\dfrac{32}{3}$$

C
$$\dfrac{8}{3}$$

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

Solution

We have,

$$16{x^2} - 9{y^2} = 144$$

$$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$$

Here, $$a=3, b=4$$

We know that the latus rectum

$$=\dfrac{2b^2}{a}$$

Therefore,

$$=\dfrac{2\times 16}{3}$$

$$=\dfrac{32}{3}$$

Hence, this is the answer.
Question 3
1 / -0

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

A
$$4$$

B
$$2$$

C
$$\dfrac {1}{5}$$

D
$$\dfrac {2}{5}$$

Solution
Question 4
1 / -0

Length of the latus rectum of the hyperbola $$xy-3x-4y+8=0$$

A
$$4$$

B
$$4\sqrt{2}$$

C
$$8$$

D
none of these

Solution

$$xy-3x-4y+8=0$$
$$x(y-3)-4(y-3)-12+8=0$$
$$(x-4)(xy-3)=4$$
$$xy=c^2$$ form
Latus nectum$$=\dfrac{2b^2}{a}=2a=2\sqrt{2}c$$
$$=4\sqrt{2}$$
Option B$$=4\sqrt{2}$$.
Question 5
1 / -0

The line $$y=mx+c$$ cut the circle $${x}^{2}+{y}^{2}={a}^{2}$$ in the distinct point $$A$$ and $$B$$. Equation of the circle having minimum radius that an be drawn through the points $$A$$ and $$B$$ is

A
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +2c\left( y-mx-c \right) =0$$

B
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +c\left( y-mx-c \right) =0$$

C
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$$

D
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ 2a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$$

Solution
Question 6
1 / -0

Length of the latusrectum of the hyperbola $$xy-3x-4y+8=0$$ is

A
$$4$$

B
$$4\sqrt{2}$$

C
$$8$$

D
$$None\,of\,these$$

Solution

Hyperbola $$=xy-3x-4y+8=0$$
(add & sub $$12$$)
$$x(y-3)-4(y-3)+8-12=0$$
$$(x-4)(y-3)=4$$
It is in form XY$$=c^2$$
where X$$=x-4$$
Y$$=y-3$$
Latus rectum $$=\dfrac{2b^2}{a}=2(a)$$ [rectangular hyperbola $$b=a$$]
We know that
$$a=\sqrt{2}.C$$
L.R$$=2\sqrt{2}\cdot C$$
$$=2.2.\sqrt{2}$$
$$=4\sqrt{2}$$.
Question 7
1 / -0

Three sides of a triangle have the equations $$L_{r} = y - m_r x - C_{r} = 0; r = 1, 2, 3$$. Then $$\lambda L_{2}L_{3} + \mu L_{3}L_{1} + \gamma L_{1}L_{2} = 0$$. where $$\lambda \neq 0, \mu \neq 0, \gamma \neq 0$$, is the equation of circumcircle of triangle if

A
$$\lambda (m_{2} + m_{3}) + \mu (m_{3} + m_{1}) + \gamma (m_{1} + m_{2}) = 0$$

B
$$\lambda (m_{2}m_{3} - 1) + \mu (m_{3}m_{1} - 1) + \gamma (m_{1}m_{2} - 1) = 0$$

C
Both $$(a)$$ and $$(b)$$ hold together

D
None of these

Solution
Question 8
1 / -0

The equation $$14x^{2}-4xy+11y^{2}-44x-58y+71=0$$ represents

A
a circle

B
an ellipse

C
a hyperbola

D
a rectangular hyperbola

Solution
Question 9
1 / -0

The equation of the circle passing through the points $$(4, 1), (6, 5)$$ and having the centre on the line $$4x+y-16=0$$ is

A
$${ x }^{ 2 }+{ y }^{ 2 }-6x-8y+15=0$$

B
$$15({ x }^{ 2 }+{ y }^{ 2 })-94x+18y+55=0$$

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

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

Solution

Given points,

$$(4,1),(6,5)$$

equation of circle $$(x-h)^2+(y-k)^2=r^2$$

$$\Rightarrow (4-h)^2+(1-k)^2=r^2$$....(1)

$$\Rightarrow (6-h)^2+(5-k)^2=r^2$$....(2)

solving the above 2 equations, we get,

$$h+2k=11$$....(3)

given, $$4h+k=16$$.....(4)

solving the above 2 equations, we get,

$$h=3,k=4$$

substituting the above values in (1), we get,

$$(4-3)^2+(1-4)^2=r^2$$

$$\therefore r=\sqrt{10}$$

Hence, the equation is,

$$(x-3)^2+(y-4)^2=(\sqrt {10})^2$$

$$x^2+y^2-6x-8y+15=0$$
Question 10
1 / -0

The length of latus rectum of the parabola $$4y^{2}+3x+3y+1=0$$ is

A
$$\dfrac {4}{3}$$

B
$$7$$

C
$$12$$

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

Solution