Conic Sections Test

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Conic Sections Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

If the tangent to the curve, $$y=x^3+ax-b$$ at the point $$(1, -5)$$ is perpendicular to the line, $$-x+y+4=0$$, then which one of the following points lies on the curve?

A
$$(-2, 2)$$

B
$$(2, -2)$$

C
$$(2, -1)$$

D
$$(-2, 1)$$

Solution

$$y=x^2+ax-b$$
$$(1, -5)$$ lies on the curve
$$\Rightarrow -5=1+a-b\Rightarrow a-b=-6$$ .(i)
Also, $$y'=3x^2+a$$
$$y'_{(1, -5)}=3+a$$ (slope of tangent)
$$\because$$ this tangent is $$\perp$$ to $$-x+y+4=0$$
$$\Rightarrow (3+a)(1)=-1$$
$$\Rightarrow a=-4$$ .(ii)
By (i) and (ii): $$a=-4$$, $$b=2$$
$$\therefore y=x^3-4x-2$$
$$(2, -2)$$ lies on this curve.
Question 2
1 / -0

If the lines $$2\mathrm{x}+3\mathrm{y}+1=0$$ and $$\mathrm{3x}- \mathrm{y}-4=0$$ lie along diameters of a circle of circumference $$ 10\pi$$, then the equation of the circle is:

A
$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}+2\mathrm{y}-23=0$$

B
$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}-2\mathrm{y}-23=0$$

C
$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}+2\mathrm{y}-23=0$$

D
$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}-2\mathrm{y}-23=0$$

Solution

Two diameters are along

$$2x+3y+1=0$$

$$3x−y−4=0$$

$$2x+3y+1=0$$ ..............$$(1)$$

$$3x−y−4=0$$ ...............$$(2) \times 3$$

Subtract $$(2)-(1)$$

$$\Rightarrow x =1$$ and $$y=-1$$

on solving we get center as $$(1,−1)$$

We know that circumference$$ =2\pi r$$

$$2\pi r=10\pi $$

$$2r=10$$

$$r=5$$

Required circle is $$(x−1)^2+(y+1)^2=52$$

$$x^2+y^2−2x+2y−23=0$$
Question 3
1 / -0

If $$5x+9=0$$ is the directrix of the hyperbola $$16{x}^{2}-9{y}^{2}=144$$, then its correponding focus is:

A
$$\left( -\cfrac { 5 }{ 3 } ,0 \right) $$

B
$$(5,0)$$

C
$$(-5,0)$$

D
$$\left( \cfrac { 5 }{ 3 } ,0 \right) $$

Solution
Question 4
1 / -0

If $${a}\neq 0$$ and the line $$2{b}{x}+3 cy+4{d}=0$$ passes through the points of intersection of the parabolas $${y}^{2}=4ax$$ and $${x}^{2}=4ay,$$ then

A
$${d}^{2}+(2{b}+3{c})^{2}=0$$

B
$${d}^{2}+(3{b}+2{c})^{2}=0$$

C
$${d}^{2}+(2{b}-3{c})^{2}=0$$

D
$${d}^{2}+(3{b}-2{c})^{2}=0$$

Solution

The equation of parabolas are $${ y }^{ 2 }=4ax$$ and $${ x }^{ 2 }=4ay$$

On solving these we get $$x=0$$ and $$x=4a$$

Also $$y=0$$ and $$y=4a$$

Therefore point of intersection of parabolas are $$A(0,0)$$ and $$B(4a,4a)$$

Also line $$2bx+3cy+4d=0$$ passes through A and B

$$\therefore d=0$$ ...(1)

$$2b.4a+3c.4a+4d=0\Rightarrow 2ab+3ac+d=0$$

$$\\ \Rightarrow a\left( 2b+3c \right) =0\Rightarrow 2b+3c=0$$ ...(2)

On squaring (1) and (2) and then adding, we get

$${ d }^{ 2 }+{ \left( 2b+3c \right) }^{ 2 }=0$$
Question 5
1 / -0

The circle $$x^{2}+y^{2}-8x=0$$ and hyperbola $$\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$$ intersect at the points $$A$$ and $$B$$.
then the equation of the circle with $$AB$$ as its diameter is

A
$$x^{2}+y^{2}-12x+24=0$$

B
$$x^{2}+y^{2}+12x+24=0$$

C
$$x^{2}+y^{2}+24x-12=0$$

D
$$x^{2}+y^{2}-24x-12=0$$

Solution

Given: $$x^2+y^2-8x=0 .....(1)$$ $$\implies y^2=8x-x^2$$

$$\dfrac {x^2}{9}-\dfrac {y^2}{4}\Rightarrow 1$$

$$\Rightarrow\dfrac {x^2}{9}-\dfrac {(8x-x^2)}{4}=1$$

$$\Rightarrow4x^2-9(8x-x^2)=36$$

$$\Rightarrow4x^2-72x+9x^2-36=0$$

$$\Rightarrow13x^2-72x-36=0$$

$$\Rightarrow x=\dfrac {72\pm \sqrt {(72)^2-4\times 13\times 36}}{26}$$

$$\Rightarrow x=6$$

Then, putting $$x=6$$ in equation (1), we get

$$\Rightarrow y=\pm \sqrt{12}=\pm \sqrt{4\times3}=\pm 2\sqrt 3$$

So, radius $$=2\sqrt 3$$

As $$AB$$ is a diameter center will be the midpoint of point $$A=(6,2\sqrt 3)$$ and $$B=(6,-2\sqrt 3)$$

By mid point formula,
Center $$=\left(\dfrac{6+6}{2},\dfrac{2\sqrt 3-2\sqrt 3}{2}\right)$$

Centre $$=(6, 0)$$

$$\Rightarrow(x-6)^2+y^2=12$$

$$\Rightarrow x^2+36-12x+y^2=12$$

$$\Rightarrow x^2-12x+y^2+24=0$$
Question 6
1 / -0

The length of the latus rectum of the parabola $$169 \left[(x-1)^2+(y-3)^2\right]=(5x-12y+17)^2$$ is:

A
$$\displaystyle \frac { 14 }{ 13 } $$

B
$$\displaystyle \frac { 12 }{ 13 } $$

C
$$\displaystyle \frac { 28 }{ 13 } $$

D
$$none\ of\ these$$

Solution

Given equation can be rewritten as
$$\displaystyle { \left( x-1 \right) }^{ 2 }+{ \left( y-3 \right) }^{ 2 }={ \left( \frac { 5x-12y+17 }{ 13 } \right) }^{ 2 }$$
$$\displaystyle \Rightarrow \quad SP=PM$$
Here, focus is $$(1, 3)$$, directrix
$$\displaystyle 5x-12y+17=0$$
$$\displaystyle \therefore $$ the distance of the focus from the directrix
$$\displaystyle =\left| \frac { 5-36+17 }{ \sqrt { 25+144 } } \right| $$
$$\displaystyle =\frac { 14 }{ 13 } =2a$$
$$\displaystyle \therefore $$ Latusrectum $$\displaystyle =2\times \frac { 14 }{ 13 } =\frac { 28 }{ 13 } $$
Question 7
1 / -0

The equation of the circle touching $$x = 0, y = 0$$ and $$x = 4$$ is

A
$$x^2 + y^2 - 4x - 4y + 16 = 0$$

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

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

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

Solution

The given circle is as shown in the following figure
It is clear that the coordinates of the centre of the circle are (2, 2) and radius = 2
Hence, the required equation of the circle is $$(x - 2)^2 + (y - 2)^2 = (2)^2$$
$$\Rightarrow x^2 + y^2 - 4x - 4y + 4 = 0$$
Question 8
1 / -0

Equation of the ellipse in its standard form is $$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

A
True

B
False

C
Nither

D
Either

Solution

Equation of ellipse in standard form is
$$\dfrac { { x }^{ 2 } }{ a^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$$
False
Question 9
1 / -0

The radius of the circle with center (0,0) and which passes through (-6,8) is

A
5

B
10

C
6

D
8

Solution

$$\displaystyle r=\sqrt{\left ( 6 \right )^{2}+\left ( -8 \right )}=10$$
Question 10
1 / -0

The equation of circle with its centre at the origin is $$x^2+y^2=r^2$$

A
True

B
False

C
Neither

D
Either

Solution

The center is a fixed point in the middle of the circle; usually given the general coordinates (h, k).

The fixed distance from the center to any point on the circle is called the radius.

A circle can be represented by two different forms of equations, the general form and the center-radius form. This discussion will focus on the center-radius form.
centre radius form
let (h,k) be the centre of a cicle and r be the radius
$${(x-h)}^{2}+{(y-k)}^{2}={r}^{2}$$
and we talk about centre at origin
than $$(h,k)$$ will become $$(0,0)$$
so, equation will become $${x}^{2}+{y}^{2}={r}^{2}$$
therefore option A will be answer.

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 9

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Conic Sections Test - 9

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SHARING IS CARING

Question 1

$$(-2, 2)$$

$$(2, -2)$$

$$(2, -1)$$

$$(-2, 1)$$

Solution

Question 2

$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}+2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}-2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}+2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}-2\mathrm{y}-23=0$$

Solution

Question 3

$$\left( -\cfrac { 5 }{ 3 } ,0 \right) $$

$$(5,0)$$

$$(-5,0)$$

$$\left( \cfrac { 5 }{ 3 } ,0 \right) $$

Solution

Question 4

$${d}^{2}+(2{b}+3{c})^{2}=0$$

$${d}^{2}+(3{b}+2{c})^{2}=0$$

$${d}^{2}+(2{b}-3{c})^{2}=0$$

$${d}^{2}+(3{b}-2{c})^{2}=0$$

Solution

Question 5

$$x^{2}+y^{2}-12x+24=0$$

$$x^{2}+y^{2}+12x+24=0$$

$$x^{2}+y^{2}+24x-12=0$$

$$x^{2}+y^{2}-24x-12=0$$

Solution

Question 6

$$\displaystyle \frac { 14 }{ 13 } $$

$$\displaystyle \frac { 12 }{ 13 } $$

$$\displaystyle \frac { 28 }{ 13 } $$

$$none\ of\ these$$

Solution

Question 7

$$x^2 + y^2 - 4x - 4y + 16 = 0$$

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

$$x^2 + y^2 + 4x + 4y + 4 = 0$$

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

Solution

Question 8

True

False

Nither

Either

Solution

Question 9

5

10

6

8

Solution

Question 10

True

False

Neither

Either

Solution

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