Hyperbola Test

Result

Hyperbola Test - 2

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 equation to the common tangents to the two hyperbolas and are

A
y = ± x ±

B
y = ± x ±

C
y = ± x ± (a² – b²)

D
y = ± x ±

Solution
Question 2
4 / -1

Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola 16y² – 9x² = 1 is

A
x² + y² = 9

B
x² + y² = 1/9

C
x² + y² = 7/144

D
x² + y² = 1/16

Solution
Question 3
4 / -1

The ellipse 4x² + 9y² = 36 and the hyperbola 4x² – y² = 4 have the same foci and they intersect at right angles then the equation of the circle through the points of intersection of two conics is

A
x² + y² = 5

B
√5 (x²+y²) – 3x – 4y = 0

C
√5 (x²+y²)+3x+4y=0

D
x² + y² = 25

Solution
Question 4
4 / -1

The equation of the common tangent to the parabola y² = 8x and the hyperbola 3x²– y² = 3 is

A
2x ± y + 1 = 0

B
x ± y + 1 = 0

C
x ± 2y + 1 = 0

D
x ± y + 2 = 0

Solution
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
None of these

Solution

Given hyperbola is 25x²−16y²=400
If (6, 2) is the midpoint of the chord, then equation of chord is T = S1
⇒25(6x)−16(2y)=25(36)−16(4)
⇒75x−16y=450−32
⇒75x−16y=418
Question 6
4 / -1

The asymptotes of the hyperbola xy–3x–2y = 0 are

A
x – 2 = 0 and y – 3 = 0

B
x – 3 = 0 and y – 2 = 0

C
x + 2 = 0 and y + 3 = 0

D
x +3 = 0 and y + 2 = 0

Solution

xy - 3x - 2y + λ = 0.
Then abc + 2fgh − af² − bg² − ch² = 0
⇒ 3/2 − λ/4 = 0
⇒ λ = 6
∴ Equation of asymptotes is xy-3x-2y+6=0
⇒ (x-2)(y-3)=0
⇒x - 2 = 0 and y - 3 = 0
Question 7
4 / -1

If the product of the perpendicular distances from any point on the hyperbola of eccentricity e = √3 on its asymptotes is equal to 6, then the length of the transverse axis of the hyperbola is

A
3

B
6

C
8

D
12

Solution

e² = (a²/ b²) + 1
3 = (a² / b²) + 1
a²/b² = 2
a² = 2b²
Product of perpendicular distance of any point on hyperbola = (a²b²)/a² + b²
= [2(b²).b²]/(2b² + b²)
= [2b²]/3 = 6
= 2b² = 18
=> b² = 9
=> b = 3
Length (2b) = 2(3)
= 6
Question 8
4 / -1

If the normal to the rectangular hyperbola xy = c² at the point `t' meets the curve again at `t₁' then t³t₁ has the value equal to

A
1

B
-1

C
0

D
None

Solution
Question 9
4 / -1

Area of triangle formed by tangent to the hyperbola xy = 16 at (16, 1) and co-ordinate axes equals

A
8

B
16

C
32

D
64

Solution

Differentiating xy=16, (xdy)/x+y=0
⇒ dy/dx=−y/x=−1/16= Slope of tangent
⇒ Its (tangent's) equation : y=−1=−1/16(x−16)
⇒ 16y−16=−x+16 ⇒ 16y=−x−32
⇒ 16y+x+32=0
It will cut x−axis at A(−32, 0)
& y−axis at B(0, −2)
⇒ Area of △OAB= 1/2×2×32
⇒ 32
Question 10
4 / -1

Locus of the middle points of the parallel chords with gradient m of the rectangular hyperbola xy = c² is

A
y + mx = 0

B
y – mx = 0

C
my – mx = 0

D
my + x = 0

Solution

Let middle point of chord is P(h,k)
Thus equation of chord with mid point P is, kx+hy=2hk
Given slope of this chord is m
⇒ −k/h = m
⇒k+mh=0
Thus locus of P(h,k) is, y+mx=0