Conic Sections Test

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Conic Sections Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The equation of the circle which bisects the circumference of the circles $$x^{2}+y^{2} =1, \;x^{2}+y^{2}-2x =3$$ and $$x^2+y^{2}+2y =3$$ is

A
$$x^{2}+y^{2}-4x+4y-1= 0$$

B
$$x^{2}+y^{2}+2x+2y-1= 0$$

C
$$x^{2}+y^{2}+2x-2y-1= 0$$

D
$$x^{2}+y^{2}-2x+4y-1= 0$$

Solution
Question 2
1 / -0

The locus of point of intersection $$P$$ of tangents to ellipse $$2x^{2}+3y^{2}=6$$ at $$A$$ and $$B$$ if $$AB$$ subtend $$90^{o}$$ angle at centre of ellipse is an ellipse whose eccentricity is equal to

A
$$\surd {5}/4$$

B
$$\surd {5}/3$$

C
$$2/\surd {5}$$

D
$$none\ of\ these$$

Solution
Question 3
1 / -0

The equation of the circle, passing through the point $$\left(2,8\right)$$, touching the lines $$4x-3y-24=0$$ and $$4x+3y-42=0$$ and having x coordinate of the centre of the circle numerically less then or equal to 8 is

A
$$x^2+y^2+4x-6y-12=0$$

B
$$x^2+y^2-4x+6y-12=0$$

C
$$x^2+y^2-4x-6y-12=0$$

D
None of these

Solution

Since $$C\left(\alpha,\beta\right)$$ be the centre of the circle.
Since the circle passes through the point $$\left(2,8\right)$$
$$\therefore$$ Radius of the circle$$=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$$
Since the circle touches the lines $$4x-3y-24=0$$ and $$4x+3y-42=0$$
$$\therefore\left|\dfrac{4\alpha-3\beta-24}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\left|\dfrac{4\alpha+3\beta-42}{\sqrt{{4}^{2}+{5}^{2}}}\right|=\sqrt{{\left(\alpha-2\right)}^{2}+{\left(\beta-8\right)}^{2}}$$ ........$$\left(1\right)$$
Solving them, we get
$$4\alpha-3\beta-24=\pm\left(4\alpha+3\beta-42\right)$$ ........$$\left(3\right)$$
$$\therefore 6\beta=18$$ or $$\beta=3$$ by taking the positive sign
and $$8\alpha=66$$ or $$\alpha=\dfrac{66}{8}=\dfrac{33}{4}$$ by taking negative sign.
Given $$\left|\alpha\right|\le 8\Rightarrow -8\le \alpha \le 8$$
$$\therefore \alpha \neq \dfrac{33}{4}$$
Put $$\beta=3$$ in equations $$\left(1\right)$$ and $$\left(3\right)$$ and equating, we get
$${\left(4\alpha-33\right)}^{2}={\left(\alpha-2\right)}^{2}+25$$
$$\Rightarrow 16{\alpha}^{2}-264\alpha+1089=25{\alpha}^{2}+725-100\alpha$$
$$\Rightarrow 9{\alpha}^{2}+164\alpha-364=0$$ which is quadratic in $$\alpha$$
$$\therefore \alpha=\dfrac{-164\pm\sqrt{{164}^{2}-36\times -364}}{2\times 9}=\dfrac{-164\pm\sqrt{{164}^{2}+36\times 364}}{18}$$
$$=\dfrac{-164\pm\sqrt{40000}}{18}=\dfrac{-164\pm 200}{18}=2,\dfrac{-182}{9}$$
But $$-8\le \alpha\le 8$$ $$\therefore \alpha=2$$
Now $${r}^{2}={\left(\alpha-2\right)}^{2}+{\left(3-8\right)}^{2}={\left(2-2\right)}^{2}+{\left(3-8\right)}^{2}=25$$ where $$r$$ is the radius of the circle
Hence the required equation of the required circle is
$${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=25$$
$${x}^{2}+{y}^{2}-4x-6y+4+9-25=0$$ or $${x}^{2}+{y}^{2}-4x-6y-12=0$$
Question 4
1 / -0

A circle has ccentre $$C$$ on axes of parabola and it touches the parabola at point $$P$$. $$CP$$ makes an angle of $$120^{o}$$ with axis of parabola. If radius of circle is $$2$$, then latus rectum of parabola is

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution
Question 5
1 / -0

If $${ e }_{ 1 }$$ is the eccentricity of the ellipse $$\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1and{ \quad e }_{ 2 }$$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $${ e }_{ 1 }{ e }_{ 2 }=1$, then equation of the hyperbola is ___________________.

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

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

C
$$\dfrac { { x }^{ 2 } }{ 9 } -\dfrac { { y }^{ 2 } }{ 25 } =1$$

D
none of these

Solution
Question 6
1 / -0

Length of latus rectum of the parabola $$9x^{2}+16y^{2}+24xy-4x+3y=0$$ is :

A
$$\dfrac{1}{20}$$

B
$$\dfrac{1}{4}$$

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

D
$$1$$

Solution
Question 7
1 / -0

Find the equation of the circle having $$(1, -2)$$ as its centre and passing through the intersection of the lines $$3x+y=14$$ and $$2x+5y=18$$.

A
$$x^2-y^2+2x+4y+20=0$$

B
$$x^2+y^2+2x-4y+20=0$$

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

D
$$x^2-y^2+2x-4y+20=0$$

Solution
Question 8
1 / -0

The equation of the circle passing through $$(4,6)$$ and having centre $$(1,2)$$ is

A
$${x}^{2}+{y}^{2}-2x-4y-20=0$$

B
$${x}^{2}+{y}^{2}-2x+4y-20=0$$

C
$${x}^{2}+{y}^{2}+2x-4y-20=0$$

D
$${x}^{2}+{y}^{2}+2x+4y-20=0$$

Solution

Let coordinates of point A are $$\left( 4,6 \right) $$
As circle is passing through point A, point A lies on circle.

Let center of circle is $$C\left( 1,2 \right) $$
$$\therefore h=1$$ and $$k=2$$

Thus, AC is radius of circle.

By distance formula,
$$AC=r=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 6-2 \right) }^{ 2 } } $$

$$\therefore r=\sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 } } $$

$$\therefore r=\sqrt { 9+16 } $$

$$\therefore r=\sqrt { 25 } $$

$$\therefore r=5$$

Thus, equation of circle is,
$${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$$

$${ \left( x-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }=\left( 5 \right) ^{ 2 }$$

$${ x }^{ 2 }-2x+1+y^{ 2 }-4y+4=25$$

$$\therefore { x }^{ 2 }+y^{ 2 }-2x-4y-20=0$$

Thus, answer is option (A)
Question 9
1 / -0

If the latus rectum subtends a right angel at the center of a hyperbola, then its eccentricity is

A
$$\sqrt{2}$$

B
$$\dfrac{\sqrt{3}+1}{2}$$

C
$$\dfrac{\sqrt{5}-1}{2}$$

D
$$\dfrac{\sqrt{3}+\sqrt{5}}{2}$$

Solution
Question 10
1 / -0

If the length of the latus rectum of the parabola $$289(x-3)^{2}+289(y-1)^{2}=(15x-8y+13)^{2}$$ is $$k$$, then $$[k]=?$$

A
$$5$$

B
$$2$$

C
$$3$$

D
$$1$$

Solution