Conic Sections Test

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Conic Sections Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Eccentricity of the hyperbola $$x^{2}-y^{2}=4$$ is

A
$$2$$

B
$$1$$

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

D
$$\sqrt {2}$$

Solution
Question 2
1 / -0

A circle has its centre on the $$y-axis$$ and passes through the origin, touches another circle with centre $$(2,2)$$ and radius 2, then the radius of the circle is

A
$$1$$

B
$$1/2$$

C
$$1/3$$

D
$$1/4$$

Solution
Question 3
1 / -0

$$L L ^ { '}$$ is the latus rectum of an ellipse and $$\triangle S ^ { \prime } L L ^ { ' }$$ is an equilateral triangle. Then $$e =$$

A
$$\dfrac { 1 } { \sqrt { 2 } }$$

B
$$\dfrac { 1 } { \sqrt { 3 } }$$

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

D
$$\dfrac { 1 } { \sqrt { 7 } }$$

Solution
Question 4
1 / -0

Length of the latus rectum of the parabola $$\sqrt {x}+\sqrt {y}=\sqrt {a}$$ is

A
$$a\sqrt {2}$$

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

C
$$a$$

D
$$2a$$

Solution

Given that,

$$\sqrt {x} + \sqrt {y} = \sqrt {a}$$

$$\dfrac {\sqrt {x}}{\sqrt {a}} + \dfrac {\sqrt {y}}{\sqrt {a}} = 1$$

$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{a}} = 1 ..... (1)$$

But we know that

$$\sqrt {\dfrac {x}{a}} + \sqrt {\dfrac {y}{b}} = 1$$

is a parabola with focus

$$F = \dfrac {ab}{c^{3}}$$ where $$c$$

Directrix $$ax + by = 0$$ Since the semi lotus rectum of a parabola is the distance from focus to directrix, we have

Latus rectum $$= \dfrac {2. a\left (\dfrac {ab^{2}}{c^{2}}\right ) + b\left (\dfrac {a^{2}b}{c^{2}}\right )}{\sqrt {a^{2} + b^{2}}}$$

$$= \dfrac {4a^{2}b^{2}}{c^{3}}$$

From equation (1) to

$$a = b$$ so,

Latus rectum $$= \dfrac {4a^{2} (a^{2})}{(\sqrt {a^{2} + a^{2}})^{3}}$$

$$= \dfrac {4a^{4}}{(\sqrt {2a^{2}})^{3}}$$

$$= \dfrac {4a^{4}}{(\sqrt {2}a)^{3}}$$

$$= \sqrt {2}a$$

Latus rectum $$= \sqrt {2}a$$

Hence, this is the answer.
Question 5
1 / -0

If there is exactly one tangent at a distance of $$4$$ units from one of the focus of $$\dfrac {x^{2}}{a^{2}}+\dfrac {y^{2}}{a^{2}-16}=1,a > 4$$, the length of latus rectum is :-

A
$$16$$

B
$$\dfrac {8}{3}$$

C
$$12$$

D
$$15$$
Question 6
1 / -0

Distance between the foci of the curve represented by the equation $$x=3+4\cos\theta, y=2+3\sin\theta$$, is?

A
$$3\sqrt{7}$$

B
$$2\sqrt{7}$$

C
$$\sqrt{7}$$

D
$$\dfrac{\sqrt{7}}{2}$$

Solution
Question 7
1 / -0

The equation $$\dfrac{x^2}{2-r}+\dfrac{y^2}{r-5}+1=0$$ represents an ellipse, if

A
$$r>2$$

B
$$r\in \left(2,\:\dfrac{7}{2}\right)\cup \left(\dfrac{7}{2},5\right)$$

C
$$r>5$$

D
$$r<2$$

Solution

Equating the equation of the ellipse with the second-degree equation
$$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$$ with $$\dfrac{{x}^{2}}{2-r}+\dfrac{{y}^{2}}{r-5}+1=0$$
we get $$A=\dfrac{1}{2-r}, B=0, C=\dfrac{1}{r-5},D=0,E=0$$ and $$F=1$$
For the second degree equation to represent an ellipse, the coefficients must satisfy the discriminant condition $${B}^{2}-4AC<0$$ and also $$A\neq C$$
$$\Rightarrow {\left(0\right)}^{2}-4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$$
$$\Rightarrow -4\left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)<0$$
$$\Rightarrow \left(\dfrac{1}{2-r}\right)\left(\dfrac{1}{r-5}\right)>0$$
$$\Rightarrow \left(2-r\right)\left(r-5\right)<0$$
$$\Rightarrow \left(r-2\right)\left(r-5\right)>0$$
$$\Rightarrow r>2$$
Question 8
1 / -0

A variable circle is drawn to touch the x-axis at the origin.The locus of the pole at the straight line $$6 x + m y + n = 0$$ w.r.t. the variable circle has the equation:-

A
$$x ( m y - n ) - e y ^ { 2 } = 0$$

B
$$x ( m y + n ) - e y ^ { 2 } = 0$$

C
$$x ( m y - n ) + \ell y ^ { 2 } = 0$$

D
none

Solution
Question 9
1 / -0

General solution of the equation $$ y=x\dfrac{dy}{dx}+\dfrac {dx}{dy}$$ represents _____________.

A
a straight line or hyperbola

B
a straight line or parabola

C
a parabola or hyperbola

D
circles

Solution
Question 10
1 / -0

Equation of circles which touch both the axes and whose centres are at a distance of $$2\sqrt {2}$$ units from origin are

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

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

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

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

Solution

Since $$h=k$$ (radius )
$$\sqrt{h^{2}+k^{2}}=2\sqrt{2}$$
$$h^{2}+k^{2}=8$$
$$2h^{2}=8$$ (Taking circle $$1^{st}$$ Quad)
$$h=2$$
$$\therefore k=2$$ in
$$\therefore$$ center $$=(2,2)$$
Eqn. of circle having centre
$$(h,k)$$ & radius $$=r$$
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
$$\therefore (x-2)^{2}+ (y-2)^{2}=4$$