Conic Sections Test

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Conic Sections Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

If the curves $$f(x) = e^{x}$$ and $$g(x) = kx^{2}$$ touches each other then the value of $$k$$ is equal to

A
$$2$$

B
$$\dfrac {e^{2}}{4}$$

C
$$\dfrac {e^{2}}{2}$$

D
$$\dfrac {e}{2}$$

Solution
Question 2
1 / -0

Find the equation of the circle which passes through the point $$(1, 1)$$ & which touches the $$x^2 + y^2 + 4x - 6y - 3 = 0$$ at the point $$(2, 3)$$ on it.

A
$$x^2 + y^2 + x - 6y + 3 = 0$$

B
$$x^2 + y^2 + x - 6y - 3 = 0$$

C
$$x^2 + y^2 + x + 6y + 3 = 0$$

D
$$x^2 + y^2 + x - 3y + 3 = 0$$

Solution

Correction(in 3rd last step): $$x^2+y^2 ``+"x-6y=\dfrac{9}{4}-\dfrac{1}{4}+4-9$$
Change $$-x$$ to $$+x$$ in following steps
Question 3
1 / -0

Let circles $${C}_{1}$$ and $${C}_{2}$$ an Argand plane be given by $$\left| z+1 \right| =3$$ and $$\left| z-2 \right| =7\ \ $$ respectively. If a variable circle $$\left| z-{ z }_{ 0 } \right| =r\quad $$ be inside circle $${C}_{2}$$ such that it touches $${C}_{1}$$ externally and $${C}_{2}$$ internally then locus of $${z}_{0}$$ describes a conic $$E$$ whose eccentricity is equal to

A
$$\cfrac { 1 }{ 10 } $$

B
$$\cfrac { 3 }{ 10 } $$

C
$$\cfrac { 5 }{ 10 } $$

D
$$\cfrac { 7 }{ 10 } $$

Solution

We have $${ C }_{ 1 }:{ \left( x+1 \right) }^{ 2 }+{ y }^{ 2 }=9$$
$${ C }_{ 2 }:{ \left( x-2 \right) }^{ 2 }+{ y }^{ 2 }=49$$
Now $$C{ C }_{ 1 }=r+{ r }_{ 1 }$$
and $$C{ C }_{ 2 }={ r }_{ 2 }-r$$
$$\Rightarrow C{ C }_{ 1 }+C{ C }_{ 2 }={ r }_{ 1 }+{ r }_{ 2 }$$
Locus of C is an ellipse with focus at
$${C}_{1}$$ and $${C}_{2}$$
Now $${ r }_{ 1 }+{ r }_{ 2 }=2a=10....(1)\quad $$
and $${ d }_{ { C }_{ 1 }{ C }_{ 2 } }(focal\quad length)=2ae=3....(2)$$
(1) and (2) $$\Rightarrow$$ eccentricity $$e$$ is $$\cfrac { 3 }{ 10 } $$
Question 4
1 / -0

Find the locus of the point of intersection of the lines $$\sqrt{3}x-y-4\sqrt{3} \lambda=0$$ and $$\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$$ for different values of $$\lambda$$.

A
$$4x^2-y^2=48$$

B
$$x^2-4y^2=48$$

C
$$3x^2-y^2=48$$

D
$$y^2-3x^2=48$$

Solution

Let $$(h,k)$$ be the point of intersection of the given lines. Then,

$$\sqrt{3}h-k-4\sqrt{3} \lambda=0$$ and $$\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$$

$$\sqrt{3}h-k=4\sqrt{3}\lambda$$ and $$\lambda(\sqrt{3}h+k)=4\sqrt{3}$$

$$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$$

$$3h^2-k^2=48$$

Hence, the locus of (h,k) is $$3x^2-y^2=48$$.
Question 5
1 / -0

Find the equation of the circle which passes through the point (1, 1) & which touches the circle $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$ at the point $$(2, 3)$$ on it.

A
$$x^{2} + y^{2} + x - 6y + 3 = 0$$

B
$$x^{2} + y^{2} + x - 6y - 3 = 0$$

C
$$x^{2} + y^{2} + x + 6y + 3 = 0$$

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

Solution

REF.Image.
Let's consider given circle

$$ S_{1} = 0 $$ &

equation of required circle

$$as\, S_{2} =0 $$

$$ Eq^n $$ of tangent at P(2,3) to $$ S_{1} = 0 $$

is $$ x = 2 $$

(because equation of CP is y = 3 and tangent at P is $$ \perp ^{le}$$ to
Required circle should touch $$ S_{1} =0 $$ at $$P(2,3)$$

$$ \Rightarrow $$ tangent at $$ P(2,3) $$ to $$ S_{1} = 0$$ should be
common tangent for $$ S_{1} = 0 \, \&\, S_{2} = 0$$

$$ \Rightarrow $$ Required circle should touch the line x = 2
at P(2,3)

$$ eq^n $$ of $$ S_{2} =0 $$ will belong to family of circles
$$ S+\lambda L = 0$$ [ S is point circle of L is tangent at p]
i.e.$$ (x-2)^2 + (y-3)^2 +\lambda (x-2) = 0$$

It should satisfy (1,1) $$ \Rightarrow (1-2)^2+(1-3)^2+ \lambda (1-2) = 0$$
$$ \Rightarrow \lambda = 5 $$

equation of required circle
$$ (x-2)^2 + (y-3)^2 + 5(x-2) = 0$$
$$ \Rightarrow x^2 +y^2 +x -6y +3 =0 $$
Question 6
1 / -0

The equation of the hyperbola whose foci are $$(6, 5), (-4, 5)$$ and eccentricity $$5/4$$ is?

A
$$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$$

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

C
$$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$$

D
$$\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$$

Solution

Let the centre of hyperbola be $$(\alpha , \beta)$$
As $$y=5$$ line has the foci, it also has the major axis.
$$\therefore \dfrac{(x-\alpha)^2}{a^2}-\dfrac{(y-\beta)^2}{b^2}=1$$
Midpoint of foci = centre of hyperbola
$$\therefore \alpha =1, \beta =5$$
Given, $$e=\dfrac{5}{4}$$.
We know that foci is given by $$(\alpha \pm ae, \beta)$$
$$\therefore \alpha +ae=6$$
$$\Rightarrow 1+\dfrac{5}{4}a=6\Rightarrow a=4$$
Using $$b^2=a^2(e^2-1)$$
$$\Rightarrow b^2=16\left(\dfrac{25}{16}-1\right)=9$$
$$\boxed{\therefore\ Equation\ of\ hyperbola \Rightarrow\\ \dfrac{(x-1)^2}{16}-\dfrac{(y-5)^2}{9}=1}$$
Question 7
1 / -0

If the curve y = | x- 3| touches the parabola $$y^2 = \lambda (x-4), \lambda >0$$, then latus rectum of the parabola, is

A
2

B
4

C
8

D
16

Solution

$$y^2=\lambda (x-4)$$

The basic equation of parabola:

$$(y-k)^2=4a(x-h)$$

$$\Rightarrow (h,k)=(4,0)$$

As latus rectum of parabola is given by the equation: $$|4a|$$

$$\Rightarrow 4a=\lambda $$

Considering $$a=1\Rightarrow \lambda =4$$

$$\therefore \lambda =4$$
Question 8
1 / -0

The equation of the circle passing through $$(4,\ 5)$$ having the centre at $$(2 ,\ 2)$$ is

A
$$x^{2} + y^{2} + 4x + 4y-5=0$$

B
$$x^{2} + y^{2} - 4x - 4y-5=0$$

C
$$x^{2} + y^{2} -4x = 13$$

D
$$x^{2} + y^{2} - 4x - 4y + 5=0$$

Solution

Let the equation of circle be
$$x^{2}+y^{2}+2gx+2fy+c=0$$
$$\therefore$$ as centre is $$(-g, -f)$$
$$g=-2, f=-2$$ ($$\therefore$$ as centre if given $$(2,2)$$)
$$\therefore$$ eqn is $$x^{2}+y^{2}-4x-4y+c=0$$
Also circle passes through $$(4,5)$$
$$\therefore$$ it satisfies $$(4,5)$$
$$\therefore (4)^{2}+(5)^{2}-4(4)-4(5)+C=0$$
$$\therefore 16+25-16-20+C=0$$
$$\therefore 5+C=0$$
$$\therefore C=-5$$
Hence eqn of circle is
$$x^{2}+y^{2}-4x-4y-5=0$$
Question 9
1 / -0

state whether following statements are true or false
Statement 1 : $$\sqrt {(x-1)^2 +y^2} + \sqrt{(x+1)^2 + y^2} = 4$$ represent equation of ellipse
Statement 2 : The locus of point which moves such that sum of its distance from two fixed points is constant is an ellipse.

A
Statement 1 is true, statement 2 is true and statement 2 is correct explanation for statement 1

B
Statement 1 is true, statement 2 is true and statement 2 is NOT correct explanation for statement 1

C
Statement 1 is true, statement 2 is false

D
Statement 1 is false, statement 2 is true

Solution
Question 10
1 / -0

Directions For Questions

If $$7{l}^{2}-9{m}^{2}+8l+1=0$$ and we have to find the equation of circle having $$lx+my+1=0$$ is a tangent and we can adjust given condition as $$16{l}^{2}+8l+1=9\left({l}^{2}+{m}^{2}\right)$$
or $${\left(4l+1\right)}^{2}=9\left({l}^{2}+{m}^{2}\right)\Rightarrow \dfrac{\left|4l+1\right|}{\sqrt{\left({l}^{2}+{m}^{2}\right)}}=3$$
Center of circle$$=\left(4,0\right)$$ and radius$$=3$$ when any two non-parallel lines touching a circle, then centre of circle lies on angle bisector of lines.

...view full instructions

If $$16{m}^{2}-8l-1=0,$$ then equation of the circle having $$lx+my+1=0$$ is a tangent is

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

B
$${x}^{2}+{y}^{2}-8x=0$$

C
$${x}^{2}+{y}^{2}+8y=0$$

D
$${x}^{2}+{y}^{2}-8y=0$$

Solution

$$\because 16{m}^{2}=8l+1$$
$$\Rightarrow 16\left({l}^{2}+{m}^{2}\right)=16{l}^{2}+8l+1={\left(4l+1\right)}^{2}$$
or $$4\sqrt{\left({l}^{2}+{m}^{2}\right)}=\left|4l+1\right|$$
or $$\dfrac{\left|4l+1\right|}{\sqrt{\left({l}^{2}+{m}^{2}\right)}}=4$$
$$\therefore$$ Centre$$\equiv\left(4,0\right)$$ and radius$$\equiv4$$
Equation of circle is $${\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}={4}^{2}$$
$$\Rightarrow {x}^{2}+16-8x+{y}^{2}=16$$
or $$ {x}^{2}-8x+{y}^{2}=0$$ is the required equation of the circle

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Conic Sections Test - 30

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Conic Sections Test - 30

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SHARING IS CARING

Question 1

$$2$$

$$\dfrac {e^{2}}{4}$$

$$\dfrac {e^{2}}{2}$$

$$\dfrac {e}{2}$$

Solution

Question 2

$$x^2 + y^2 + x - 6y + 3 = 0$$

$$x^2 + y^2 + x - 6y - 3 = 0$$

$$x^2 + y^2 + x + 6y + 3 = 0$$

$$x^2 + y^2 + x - 3y + 3 = 0$$

Solution

Question 3

$$\cfrac { 1 }{ 10 } $$

$$\cfrac { 3 }{ 10 } $$

$$\cfrac { 5 }{ 10 } $$

$$\cfrac { 7 }{ 10 } $$

Solution

Question 4

$$4x^2-y^2=48$$

$$x^2-4y^2=48$$

$$3x^2-y^2=48$$

$$y^2-3x^2=48$$

Solution

Question 5

$$x^{2} + y^{2} + x - 6y + 3 = 0$$

$$x^{2} + y^{2} + x - 6y - 3 = 0$$

$$x^{2} + y^{2} + x + 6y + 3 = 0$$

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

Solution

Question 6

$$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1$$

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

$$\displaystyle\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1$$

$$\displaystyle\frac{(x-1)^2}{4}-\frac{(y-5)^2}{9}=1$$

Solution

Question 7

2

4

8

16

Solution

Question 8

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

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

$$x^{2} + y^{2} -4x = 13$$

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

Solution

Question 9

Statement 1 is true, statement 2 is true and statement 2 is correct explanation for statement 1

Statement 1 is true, statement 2 is true and statement 2 is NOT correct explanation for statement 1

Statement 1 is true, statement 2 is false

Statement 1 is false, statement 2 is true

Solution

Question 10

$${x}^{2}+{y}^{2}+8x=0$$

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

$${x}^{2}+{y}^{2}+8y=0$$

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

Solution

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