Hyperbola Test

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Hyperbola Test - 5

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Weekly Quiz Competition

Question 1
4 / -1

If the equation 4x² + ky² = 18 represents a rectangular hyperbola, then k =

A
4

B
–4

C
3

D
none of these

Solution

We know that the equation
ax² + by² + 2hxy + 2gx + 2fy + c = 0
represent a rectangular hyperbola if Δ ≠ 0, h² > ab and a + b = 0.
∴ The given equation represents a rectangular hyperbola if 4 + k = 0 i.e. k = -4
Question 2
4 / -1

The locus of the point of intersection of tangents drawn at the extremities of normal chords to hyperbola xy = c²

A
(x² - y²)² + 4c²xy = 0

B
(x² + y²)² + 4c²xy = 0

C
(x² - y²)² + 4c xy = 0

D
(x² + y²)² + 4cxy = 0

Solution

Polar is xy₁ + x₁y = 2c².. (1)
Let normal chord at (h,k) be
hx - ky = h² - k²...... (2)
From 1 and
And hk = c² eliminate h, k and λ
Question 3
4 / -1

The asymptotes of the hyperbola hx + ky = xy are

A
x - k = 0, y - h = 0

B
x + h = 0, y + k = 0

C
x - h = 0, y - k = 0

D
x + k = 0, y + k = 0

Solution

hx + ky - xy = 0
let asymptotes be hx + ky - xy + c = 0 ; This represents a pair of straight lines if Δ = 0
i.e. c =-hk ∴ asymptotes are xh+yk-xy-hk =0 ⇒ (x - k) (y - h) = 0
Question 4
4 / -1

The equation of a line passing through the centre of a rectangular hyperbola is x – y –1 = 0. if one of its asymptote is 3x-4y-6 = 0, then equation of its other asymptote is

A
4x + 3y + 17 = 0

B
4x + 3y -17 = 0

C
4x + 3y -15 = 0

D
4x + 3y + 15 = 0

Solution

Pt of intersection of x-y-1=0 and 3x-4y-6=0 is (-2,-3) other asymptote will be in the form of 4x + 3y + λ =0 and it should pass through (-2,-3). Thus λ = -17
Question 5
4 / -1

Tangents are drawn to 3x²- 2y² = 6 from a point P. If these tangents intersects the coordinate axes at concyclic points, The locus of P is

A
x² + y² = 5

B

C
x² - y² = 13

D

Solution

x²/2 - y²/3 = 1
= [y² + (3)2]/[x² - (2)²] = 1
=> [y² + 9]/[x² - 4] = 1
=> [y² + 9] = [x² - 4]
x² - y² = 13
Question 6
4 / -1

If the curve cut each other orthogonally then

A
a² b² = α² + β²

B
a² - b² = α² - β²

C
a² - b² = α² + β²

D
a² + b² = α² - β²

Solution

Slope of

------(1)
now solving the curves

-------(2)
from (1) & (2)
Question 7
4 / -1

Let a and b be non–zero real numbers. Then, the equation (ax² + by² + (C) (x² – 5xy + 6y²) = 0 represents

A
four straight lines, when c = 0 and a, b are of the same sign

B
two straight lines and a circle, when a = b, and c is of sign opposite to that of a

C
two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

D
a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a.

Solution

x² – 5xy + 6y² = 0 represent a pair of lines passing through origin
ax² + ay² = - c

⇒ ax² + ay² + c = 0 represent a circle
Question 8
4 / -1

The asymptotes of the curve 2x² + 5xy + 2y² + 4x + 5y = 0 are given by

A
2x² + 5xy + 2y² + 4x + 5y + 2 = 0

B
2x² + 5xy + 2y² + 4x + 5y - 2 = 0

C
2x² + 5xy + 2y² + 4x + 5y - 4 = 0

D
None of these

Solution

The hyperbola is given by
2x² + 5xy + 2y² + 4x + 5y = 0 ............................(i)
Since the equation of hyperbola will differ from equation of asymptote by a constant. So, equation of asymptote is
2x² + 5xy + 2y³ + 4x + 5y + λ = 0 ......................(ii)
If (ii) represents 2 straight lines, we must have
abc + 2fgh - af²- bg² - ch² = 0
or 22λ + 2.5 / 2.4 / 2.5 / 2 - 2.25 / 4 - 2.4 - λ (25 / 4) = 0
or 9λ = 18 ∴ λ = 2
2x² + 5xy + 2y³ + 4x + 5y + 2 = 0
Question 9
4 / -1

A hyperbola passing through origin has 3x – 4y – 1 = 0 and 4x – 3y – 6 = 0 as its asymptotes. Then the equation of its transverse axis is

A
x – y – 5 = 0

B
x + y + 1 = 0

C
x + y – 5 = 0

D
x – y – 1 = 0

Solution

Asymptotes are equally inclined to the axes of hyperbola Find the bisector of the asymptotes which bisects the angle containing the origin.
Question 10
4 / -1

If the vertex of a hyperbola bisects the distance between its centere and the corresponding focus, then ratio of square of its conjugate axis of the square of its transverse axis is

A
2

B
4

C
6

D
3

Solution

a = ae, i.e.,