Hyperbola Test

Result

Hyperbola Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

Equation of the common tangent to y² = 8x and 3x² - y² = 3 is

A
2x - y +1 = 0

B
x + 2y - 5 = 0

C
2x + y - 6 = 0

D
2x - y + 7 = 0

Solution

a = 2 and c² = a²m² - b²
Question 2
4 / -1

The equations of the common tangents to the two hyperbolas and

A

B

C

D
Question 3
4 / -1

If the tangent at the point (h, k) on the hyperbola meets the auxiallary circle of the hyperbola in two points whose ordinates y₁, y₂ then

A
1/k

B

C

D
2/k

Solution

Equation of the tangent at (h, k) is

Equation of auxiallary circle x² + y² = a²
Question 4
4 / -1

If x = 9 is the chord of contact of the hyperbola x² - y² = 9, then the equation of the tangent at one of the points of contact is

A

B

C

D

Solution

Solve x = 9 and x² – y² = 9 equation of tangent at point of contact is S₁ = 0
Question 5
4 / -1

Equation of the chord of the hyperbola 25x² - 16y² = 400, which is bisected at the point (6, 2) is

A
16x - 75y = 418

B
75x -16y = 418

C
25x - 4y = 400

D
4y + 25x = 100

Solution

Equation of the chord S₁ = S₁₁
Question 6
4 / -1

The mid-point of the chord 4x – 3y = 5 of the hyperbola 2x² - 3y² = 12 is

A

B
(2,1)

C

D

Solution

Given, 4x - 3y = 5 and 2x² - 3y² = 12
Question 7
4 / -1

The equation of the transverse and conjugate axes of a hyperbola are respectively x + 2y – 3 = 0, 2x – y + 4 = 0 and their respective lengths are √2 and 2/√3. The equation of the hyperbola is

A

B

C
2(2x - y + 4)² -3 (x + 2y - 3)² = 1

D
2(x + 2y -3)² -3 (2x - y + 4)² = 1

Solution

Equation of the hyperbola is

Where a₁x + b₁y + c₁ = 0, b₁x - a₁y + c₂ = 0 are conjugate and transverse axes respectively and a, b are lengths of semitransverse and semiconjugate axes respectively.
Question 8
4 / -1

If two distinct tangents can be drawn from the point (α²) on different branches of the hyperbola

A

B

C

D
α = 1
Question 9
4 / -1

A hyperbola has centre C and one focus at P(6, 8). If its two directrices are 3x + 4y + 10 = 0 and 3x + 4y -10 = 0 then CP =

A
14

B
8

C
10

D
6

Solution

is nearest to 3x + 4y -10 = 0

CP = ae = 10
Question 10
4 / -1

A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x² + 4y² = 12. Then its equation is:

A
x² cosec²θ - y² sec² θ = 1

B
x² sec² θ - y² cosec²θ = 1

C
x² sin²θ - y²cos²θ = 1

D
x² cos²θ - y² sin²θ = 1

Solution

The given ellipse is

Hence, the eccentricity e1, of the hyperbola is given by
1 = e₁ sinθ ⇒ e₁ = cosec θ ⇒ b² = sin²θ (cosec²θ - 1) = cos²θ