Mathematics Test 273

Result

Mathematics Test 273

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

The locus of a variable point whose distance from the point (2, 0) is 2/3 times its distance from the line x = 9/2 is

A
a parabola

B
a hyperbola

C
a circle

D
an ellipse

Solution
Question 2
4 / -1

The locus of a point such that two tangents drawn from it to the parabola y² = 4ax are such that the slope of one is double the other is

A
y² = 9/2 ax

B
y² = 9/4 ax

C
y² = 9ax

D
x² = 4ay

Solution
Question 3
4 / -1

T is a point on the tangent to a parabola y² = 4ax at its point P. TL and TN are the perpendiculars on the focal radius SP and the directrix of the parabola respectively. Then

A
SL = 2 (TN)

B
3 (SL) = 2 (TN)

C
SL = TN

D
2 (SL) = 3 (TN)

Solution
Question 4
4 / -1

Tangents are drawn from the points on the line x – y + 3 = 0 to parabola y² = 8x. Then the variable chords of contact pass through a fixed point whose coordinates are

A
(3, 2)

B
(2, 4)

C
(3, 4)

D
(4, 1)

Solution

Let (k,k+3) be the point on the line x−y+3=0

Equation of chord of contact is S₁=0

⇒yy1=4(x+x1)

⇒y(k+3)=4(x+k)

⇒4x−3y−k(y−4)=0

Therefore, straight line passes through fixed point (3,4)
Question 5
4 / -1

The focus of the parabola x²−8x+2y+7 = 0 is

A
4, (9/2)

B

C

D
(4, 4)

Solution

Parabola is x² – 8x + 2y + 7 = 0

∴ (x – 4)²= – 2y – 7 + 16

∴ (x – 4)² = – 2[y – (9/2)]

∴ x² = – 4ay

⇒ x = x – 4, y = y – (9/2) and 2 = 4a

i.e. a = (1/2)

Its focus is given by x = 0 and y = 0 i.e. x – 4 = 0 and y – (9/3) = 0

∴ x = 4 and y = (9/2)

∴ focus [4, (9/2)].
Question 6
4 / -1

From the focus of the parabola y² = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is

A
(x – 2)² + y² = 3

B
(x – 2)² + y² = 9

C
(x + 2)² + y² = 9

D
None

Solution

Focus of parabola y² = 8x is (2,0). Equation of circle with centre (2,0) is (x−2)² + y² = r²

Let AB is common chord and Q is mid point i.e. (1,0)

AQ² = y² = 8x

= 8×1 = 8

∴ r² = AQ² + QS²

= 8 + 1 = 9

So required circle is (x−2)² + y² = 9
Question 7
4 / -1

In an examination, ten students scored the following marks: 60, 58, 90, 51, 47, 81, 70, 95, 87, 99. The range of this data is

A
60

B
51

C
81

D
52
Solution
To find the range of the given data set, follow these steps:
1. Identify the Highest Score (Maximum):
  
  The highest score is 99.
2. Identify the Lowest Score (Minimum):
  
  The lowest score is 47.
3. Calculate the Range:
  
  Range = Maximum Score - Minimum Score
  
  Range = 99 - 47 = 52
Therefore, the range of the data is 52 marks.
Question 8
4 / -1

Two vertices of a triangle are (3,−2) and (−2, 3) and its orthocentre is (−6, 1). The coordinates of its third vertex are-

A
(1, 6)

B
(−1, 6)

C
(1, −6)

D
None of these

Solution
Question 9
4 / -1

A

B

C

D
None of the above

Solution
Question 10
4 / -1

A
143

B
147

C
137

D
157

Solution