Conic Sections Test

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Conic Sections Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The equation x² + y² = 0 represents

A
pairs of straight line.

B
an ellipse.

C
a degenerate circle.

D
an equilateral hyperbola.

Solution

The above circle can be written as (x−0)²+(y−0)² = (0)²

Here the center is (0,0) and radius is also 0 units.

So it is a degenerate circle as degenerate circle is a circle( a point) where radius is zero units.
Question 2
1 / -0

The equation(x²+y²)+5x−7y−2=0 represents

A
an empty set

B
a pair of straight lines

C
a degenerate circle

D
a circle

Solution

The general equation of the circle is x²+y²-2gh-2fy+c = 0. Sice the given equations satisfies the general equation, it represents the equation of the circle.
Question 3
1 / -0

Circumcentre of the triangle, whose vertices are (0, 0), (6, 0) and (0, 4) is

A
( 3, 2)

B
(2, 0)

C
(3, 0)

D
(0, 3)

Solution

circumcentre of a right angled triangle ABC right angled at A is b+c/2 as circumcentre of right angled triangle lies on the mid pont of the hypotenuse.

so mid point of BC=(6+0/2, 0+4/2) i.e.(3,2)
Question 4
1 / -0

If (x-a)²+ (y-b)²= c² represents a circle, then

A
b = 0

B
a = b = 0

C
c ≠ 0

D
a = 0

Solution

(x-a)²+ (y-b)²= c²here (a,b) is center and c is the radius

and radius cannot be zero because if radius is zero it will become a point or degenerate circle so c≠0.
Question 5
1 / -0

Two perpendicular tangents to the circle x² + y² = r² meet at P. The locus of P is

A
x + y = r

B
x² + y² = 2 r²

C
x² + y² = 4r²

D
x² + y² = r²/2

Solution

locus of P is a circle with centre at origin and radius √2r².This is known as the director circle of the circle x² + y² = r²

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 2

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Conic Sections Test - 2

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SHARING IS CARING

Question 1

pairs of straight line.

an ellipse.

a degenerate circle.

an equilateral hyperbola.

Solution

Question 2

an empty set

a pair of straight lines

a degenerate circle

a circle

Solution

Question 3

( 3, 2)

(2, 0)

(3, 0)

(0, 3)

Solution

Question 4

b = 0

a = b = 0

c ≠ 0

a = 0

Solution

Question 5

x + y = r

x2 + y2 = 2 r2

x2 + y2 = 4r2

x2 + y2 = r2/2

Solution

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x² + y² = 2 r²

x² + y² = 4r²

x² + y² = r²/2