Conic Sections Test

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Conic Sections Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

The centre, vertex, focus of a conic are $$(0,0),
(0,5), (0,6)$$. Its length of latus rectum is

A
$$\dfrac{11}5$$

B
$$\dfrac75$$

C
$$\dfrac{14}5$$

D
$$\dfrac{22}5$$

Solution

Given centre vertex and focus of the conic are $$C(0,0), A(0,5)$$ and $$(0,6)$$ respectively
$$\Rightarrow a=5, ae=6 \Rightarrow e=\cfrac{6}{5}$$
Thus $$b^2=a^2(e^2-1)=11$$
Therefore, length of latus rectum is $$=\cfrac{2b^2}{a}=\cfrac{22}{5}$$
Hence, option 'D' is correct
Question 2
1 / -0

The total number of real tangents that can be drawn to the ellipse $$3x^{2}+5y^{2}=32$$ and $$25x^{2}+9y^{2}=450$$ passing through $$(3,5)$$ is

A
$$0$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$(3,5)$$ lies on $$25x^2+9y^2=450$$

Therefore, one tangent can be drawn

and $$(3,5)$$ lies outside $$3x^2+5y^2=32$$ because $$S_1>0$$

Therefore, two tangents can be drawn.
So total 3 tangents
Question 3
1 / -0

The length of the latus rectum of the hyperbola $$x^{2}-4y^{2}=4$$ is

A
$$2$$

B
$$1$$

C
$$4$$

D
$$3$$

Solution

$$x^2-4y^2=4$$
$$\Rightarrow \cfrac{x^2}{2^2}-\cfrac{y^2}{1}=1$$
$$\Rightarrow a=2, b=1$$
Thus length of the latus rectum is $$=\cfrac{2b^2}{a}=1$$
Hence, option 'B' is correct
Question 4
1 / -0

A straight line is drawn through the centre of the circle $${x}^{2}+{y}^{2}=2ax$$ parallel to $${x}+2y=0$$ and intersecting the circle at $${A}$$ and $${B}$$. Then, the area of $$\Delta A{O}{B}$$ is

A
$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{5}}$$

B
$$\displaystyle \dfrac{\mathrm{a}^{3}}{\sqrt{5}}$$

C
$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{34}}$$

D
$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{3}}$$

Solution

Given equation is $$x^{2}+y^{2}=2ax$$
Centre $$=(a,0)$$ and radius $$=a$$
Therefore, equation of line having slope $$=$$ $$\dfrac {-1}{2}$$ and passing through $$(4,0)$$
$$y=\dfrac {-1}{2}(x-a)$$
Line $$AB$$: $$2y+x=a$$ -(1)
$$y=\dfrac {a+x}{2}$$
$$d=\left | \dfrac {a}{\sqrt{5}} \right |$$
So, area of $$\triangle AOB =\dfrac {1}{2}\times\ AB\ \times$$ (perpendicular distance of $$Q$$ from $$AB$$ )
$$=\dfrac {1}{2}\times 2a\times \dfrac {a}{\sqrt{5}}$$
$$=\dfrac {a^{2}}{\sqrt{5}}$$
Question 5
1 / -0

The equation $$\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x+2)^{2}+y^{2}}=5$$ represents

A
a circle

B
ellipse

C
line segment

D
an empty set

Solution

let $$A(2,0)$$, $$B(-2,0)$$ and $$P(x,y)$$ be three points
$$AB=4$$
Given that, $$\sqrt{(x-2)^{2}+y^{2}}+\sqrt{(x+2)^{2}+y^{2}}=5>AB$$
$$\Rightarrow PA+PB =$$ constant $$ >AB$$
Therefore, locus of $$P$$ is an ellipse.
Question 6
1 / -0

The length of latus rectum of the hyperbola $$4x^{2}-9y^{2}-16x-54y-101=0$$ is

A
$$\dfrac85$$

B
$$\dfrac87$$

C
$$\dfrac89$$

D
$$\dfrac83$$

Solution

$$4x^{2}-9y^{2}-16x-54y-101=0$$
$$\Rightarrow 4(x^2-4x)-9(y^2+6y)=101$$
$$\Rightarrow 4(x^2-4x+4)-9(y^2+6y+9)=101+16-81=36$$
$$\Rightarrow \cfrac{(x-2)^2}{3^2}-\cfrac{(y+3)^2}{2^2}=1$$
$$\Rightarrow a=3, b=2$$
Thus length of latus rectum of the given hyperbola is $$=\cfrac{2b^2}{a}=\cfrac{8}{3}$$
Question 7
1 / -0

The equation $$(\mathrm{x}^{2}-\mathrm{a}^{2})^{2}+(\mathrm{y}^{2}-\mathrm{b}^{2})^{2}=0$$ represent points which are

A
collinear

B
on a circle centre $$(a,b)$$

C
on a circle centre $$(0,0)$$

D
coincident

Solution

$$(x^{2}-a^{2})^{2}+(y^{2}-b^{2})^{2}= 0$$ represents points
$$x= ^{+}_{-}a$$ and $$y= ^{+}_{-}b$$
$$\therefore (a,b), (a,-b), (-a,b)$$ and $$(-a,-b)$$ are four points
These point lie on a circle whose centre is at $$(0,0)$$ and radius is $$\sqrt{a^2+b^2}$$
Question 8
1 / -0

The equation of directrix and latus rectum of a parabola are $$3x-4y+27=0$$ and $$3x-4y+2=0$$. Then the length of latus rectum is

A
$$5$$

B
$$10$$

C
$$15$$

D
$$20$$

Solution

$$d=\left | \dfrac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}} \right |$$

where $$d$$ is the distance between lines whose equations are $$ax+by+C_{1}=0$$ & $$ax+by+C_{2}=0$$

$$d=\left | \dfrac{27-2}{\sqrt{4^{2}+3^2}} \right |$$
$$=5$$
$$d=5$$
If the distance between vertex and latus rectum$$=$$distance of vertex from directrix$$=a$$
then $$d=2a=5$$
$$\Rightarrow$$ Length of latus rectum$$=4a=10$$
Question 9
1 / -0

A circle of radius $$5$$ units passes through the points $$(7,1),(9,5)$$. If the ordinate of the centre is less than $$2$$, then the equation of the circle is

A
$$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}-10\mathrm{y}+16=0$$

B
$$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}+10\mathrm{y}+16=0$$

C
$$\mathrm{x}^{2}+\mathrm{y}^{2}-24\mathrm{x}-2\mathrm{y}+120=0$$

D
$$\mathrm{x}^{2}+\mathrm{y}^{2}+24\mathrm{x}-2\mathrm{y}-120=0$$

Solution

Let $$(h, k)$$ is a centre of circle , so

$$(h-7)^2+(k-1)^2= 25$$
and $$(h-9)^2+ (k-5)^2= 25$$

$$Let x^2+y^2+2gx+2fy+c= 0$$ is circle .
so, $$(7, 1)$$ lies on circle so,

$$50+14g+2f+c= 0$$ ----(1)
$$(9, 5)$$ lies on circle

$$106+18g+10f+c= 0$$ ----(2)

and $$g^2+f^2-c= 25$$ -----(3)
From (1) & (2),
$$4g+8f+56= 0$$

$$g+2f+14= 0$$
From (1), (2), (3), (4),
$$g= -12$$ and $$f= -1, c= 120$$

so, $$eq^n$$ of circle is

$$x^2+y^2-24x-2y+120= 0$$
Question 10
1 / -0

A parabola with axis parallel to $$x$$ axis passes through $$(0, 0), (2, 1), (4, -1).$$ Its length of latus rectum is

A
$$\displaystyle \frac{2}{3}$$

B
$$\displaystyle \frac{1}{4}$$

C
$$\displaystyle \frac{7}{3}$$

D
$$\displaystyle \frac{1}{3}$$

Solution

$$(y-k)^{2}= 4a(x-h)$$
It passes through $$(0,0)$$
$$k^{2}= -4ah$$ ---(1)
It passes through $$(2,1)$$
$$(k-1)^{2}= 4a(2-h)$$ ---(2)
It passes through $$(4,-1)$$
$$(k+1)^{2}= 4a(4-h)$$ ---(3)
(2) + (3)
$$2(k^{2}+1)= 4a(6-2h)$$
$$2k^{2}+2= 24a+2k^{2}$$ ( form eq (1) )
$$4a = \dfrac{1}{3}$$
$$\therefore$$ length of LR $$= \dfrac{1}{3}$$

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 15

Result

Conic Sections Test - 15

Score

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Rank

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SHARING IS CARING

Question 1

$$\dfrac{11}5$$

$$\dfrac75$$

$$\dfrac{14}5$$

$$\dfrac{22}5$$

Solution

Question 2

$$0$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

$$2$$

$$1$$

$$4$$

$$3$$

Solution

Question 4

$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{5}}$$

$$\displaystyle \dfrac{\mathrm{a}^{3}}{\sqrt{5}}$$

$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{34}}$$

$$\displaystyle \dfrac{\mathrm{a}^{2}}{\sqrt{3}}$$

Solution

Question 5

a circle

ellipse

line segment

an empty set

Solution

Question 6

$$\dfrac85$$

$$\dfrac87$$

$$\dfrac89$$

$$\dfrac83$$

Solution

Question 7

collinear

on a circle centre $$(a,b)$$

on a circle centre $$(0,0)$$

coincident

Solution

Question 8

$$5$$

$$10$$

$$15$$

$$20$$

Solution

Question 9

$$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}-10\mathrm{y}+16=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+8\mathrm{x}+10\mathrm{y}+16=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}-24\mathrm{x}-2\mathrm{y}+120=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+24\mathrm{x}-2\mathrm{y}-120=0$$

Solution

Question 10

$$\displaystyle \frac{2}{3}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{7}{3}$$

$$\displaystyle \frac{1}{3}$$

Solution

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