Hyperbola Test

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Hyperbola Test - 1

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Question 1
4 / -1

The eccentricity of the hyperbola 4x²–9y²–8x = 32 is

A
√5/3

B

C

D
3/2

Solution

4x²−9y²−8x=32
⇒4(x²−2x)−9y2 = 32
⇒4(x²−2x+1)−9y² = 32 + 4 = 36
⇒(x−1)²]/9 − [y2]/4 = 1
⇒a²=9, b²=4
∴e=[1+b²/a²]^1/2
= [(13)^1/2]/3
Question 2
4 / -1

The locus of the point of intersection of the lines √3x - y - 4√3k = 0 and √3kx + ky - 4√3 = 0 for different values of k is

A
Ellipse

B
Parabola

C
Circle

D
Hyperbola

Solution

Given equation of line are
√3x−y−4√3k=0 …(i)
and √3kx+ky−4√3=0
From Eq. (i) 4√3–√k=3–√x−y
⇒ k=(√3x−y)/4√3
put in Eq. (ii), we get
√3x(√3x−y)/4√3)+((√3x−y)/4√3)y−4√3=0
⇒1/4(√3x²−xy)+1/4(xy−y²/√3)−4√3=0
⇒√3/4x²−y²/4√3-4√3=0
⇒3x²−y²−48=0
⇒3x²−y²=48,which is hyperbola.
Question 3
4 / -1

If the latus rectum of an hyperbola be 8 and eccentricity be 3/√5 then the equation of the hyperbola is

A
4x² – 5y² = 100

B
5x² – 4y² = 100

C
4x² + 5y² = 100

D
5x² + 4y² = 100

Solution

Give eccentricity of the hyperbola is, e= 3/(5)^1/2
⇒ (b²)/(a²) = 4/5..(1)
And latus rectum is =2(b²)/a=8
⇒ b²/a=4…(2)
By (2)/(1) a=5
∴ b²=20
Hence required hyperbola is, (x²)/25−(y²)/20=1
⇒ 4x²−5y²=100
Question 4
4 / -1

If the centre, vertex and focus of a hyperbola be (0, 0), (4, 0) and (6, 0) respectively, then the equation of the hyperbola is

A
4x² – 5y² = 8

B
4x² – 5y² = 80

C
5x² – 4y² = 80

D
5x² – 4y² = 8

Solution
Question 5
4 / -1

The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity 5/4 is

A
- = 1

B
- = 1

C
- = - 1

D
None of these

Solution

Let the centre of hyperbola be (α,β)
As y=5 line has the foci, it also has the major axis.
∴ [(x−α)²]/a² − [(y−β)²]/b² = 1
Midpoint of foci = centre of hyperbola
∴ α=1,β=5
Given, e= 5/4
We know that foci is given by (α±ae,β)
∴ α+ae=6
⇒1+(5/4a)=6
⇒ a=4
Using b² = a²(e² − 1)
⇒ b²=16((25/16)−1)=9
∴ Equation of hyperbola ⇒ [(x−1)²]/16−[(y−5)²]/9=1
Question 6
4 / -1

The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is

A

B

C

D

Solution
Question 7
4 / -1

The length of the transverse axis of a hyperbola is 7 and it passes through the point (5, –2). The equation of the hyperbola is

A

B

C

D
None of these

Solution
Question 8
4 / -1

If the eccentricity of the hyperbola x²– y² sec² a = 5 is (√3) times the eccentricity of the ellipse x² sec² a + y² = 25, then a value e of a is

A
π/6

B
π/4

C
π/3

D
π/2

Solution
Question 9
4 / -1

AB is a double ordinate of the hyperbola such that DAOB (where `O' is the origin) is an equilateral triangle, then the eccentricity e of the hyperbola satisfies

A
e >√3

B
1

C
e = 2/√3

D
e > 2/√3

Solution
Question 10
4 / -1

The equation of the tangent lines to the hyperbola x² – 2y² = 18 which are perpendicular to the line y = x are

A
y = x ± 3

B
y = –x ± 3

C
2x + 3y + 4 = 0

D
None of these

Solution

Equation of line perpendicular to x−y=0 is given by
y=−x+c
Also this line is tangent to the hyperbola x²−2y²=18
So we have m=−1, a²=18, b²=9
Thus Using condition of tangency c² = a²m²−b²
= 18−9=9
⇒ c = ±3
Hence required equation of tangent is x+y = ±3