Mathematics Test

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Mathematics Test - 1

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Weekly Quiz Competition

Question 1
3 / -1

If P(at², 2at) be one end of a focal chord of the parabola y² = 4ax, then the length of the chord is

A

B

C

D

Solution

Let end points be
Question 2
3 / -1

Two tangents are drawn from a point (-2, -1) to the curve y2 = 4x. If α is the angle between them, then |tanα| is equal to

A
3

B
√3

C

D
1/3

Solution

Combined equation of pair of tangents is given by,
SS₁=T²

⇒(y²−4x)((−1)²−4(−2))=(−1⋅y−2(x−2))²
⇒(y²−4x)(9)=(y+2x−4)²
⇒9y²−36x=y²+4x²+16−8y−16+4xy
⇒4x²−8y²+4xy+20x−8y+16=0
⇒2x²−4y²+2xy+10x−4y+8=0
Question 3
3 / -1

If 2x+3y=α, x−y=β and kx+15y=r are 3 concurrent normals of parabola y2=λx, then the value of k is

A
3

B
4

C
5

D
7

Solution

We know that the algebraic sum of slopes of all the three concurrent normals of a parabola is equal to zero.
Question 4
3 / -1

If three parabolas touch all the lines x=0, y=0 and x+y=2, then the maximum area of the triangle formed by joining their foci is

A
√3 sq. units

B
√6 sq. units

C

D

Solution

Consider a △ABC whose sides are x=0, y=0 and x+y=2

Therefore, co-ordinates of A, B & C are (0, 2), (0, 0) & (2, 0) respectively.
Since the parabolas touch all the sides, their foci must lie on the circumcircle of the Δ ABC.
We see that Δ ABC is a right angle triangle.

Now, on joining the foci of three parabolas, we get a triangle of maximum area.
Hence, foci must be the vertices of an equilateral triangle inscribed in the circumcircle.

Let side length of equilateral triangle F₁F₂F₃ be a.
Question 5
3 / -1

An equilateral triangle is inscribed in a parabola y2 = 4ax, where one vertex is

A
8√3a

B
2√3a

C
3√2a

D
√3a

Solution

The given parabola is y2=4ax ...(i)
Let OA(=l) be the side of equilateral triangle.
Then OL=lcos30°= √3l/2
and LA=lsin 30°= l/2

⇒ l=8√3a
Hence the length of the side of the triangle = 8√3a units.
Question 6
3 / -1

The length of the latus rectum and equation of the directrix of the parabola y²=−8x

A
8, x =2

B
9, y=x+2

C
7, x=2

D
6, x=y+2

Solution

Given, y²=−8x
Length of the latus rectum = 4a = 8
⇒a=2
Hence, the equation of the directrix is x=2
Question 7
3 / -1

The mirror image of the directrix of the parabola y²=4(x+1) in the line mirror x+2y=3, is

A
x=−2

B
4y−3x=16

C
x−3y=0

D
x+y=0

Solution

for the given parabola , y²=4(x+1)

directrix is x=−2. and Any point on it is (−2, k)

let mirror image of (-2,k) in the line x+2y=3 is (x,y)

⇒ 4y -3x = 16 is the equation of the mirror image of the directrix.
Question 8
3 / -1

The shortest distance (in units) between the parabolas y² = 4x and y²=2x−6 is

A
2

B
√5

C
3

D
none of these

Solution

Shortest distance between two curves occurs along the common normal.
Normal to y²=4x at (m², 2m) is
y+mx−2m−m³=0

⇒m=0, ±2
So, points will be (4, 4) and (5, 2) or (4,−4) and (5,−2)
Hence, shortest distance will be
Question 9
3 / -1

(3 + w + 3w²)⁴ equals

A
16

B
16w

C
16 w²

D
none of these

Solution

[3 (1 + w²)] + w]⁴ = (2w)⁴ = 16w⁴ = 16w
Question 10
3 / -1

The sum of 40 terms of an A.P. whose first term is 2 and common difference 4, will be

A
1600

B
3200

C
200

D
2800

Solution