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Test: Hyperbola- 2 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Hyperbola- 2

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Test: Hyperbola- 2 - Question 1

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

Detailed Solution for Test: Hyperbola- 2 - Question 1

 a = 2 and c2 = a2m2 - b2

Test: Hyperbola- 2 - Question 2

The equations of the common tangents to the two hyperbolas  and 

Test: Hyperbola- 2 - Question 3

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 

Detailed Solution for Test: Hyperbola- 2 - Question 3

Equation of the tangent at (h, k) is 

Equation of auxiallary circle x2 + y2 = a2

Test: Hyperbola- 2 - Question 4

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 

Detailed Solution for Test: Hyperbola- 2 - Question 4

Solve x = 9 and x2 – y2 = 9 equation of tangent at point of contact is S1 = 0

Test: Hyperbola- 2 - Question 5

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

Detailed Solution for Test: Hyperbola- 2 - Question 5

Equation of the chord S1 = S11

Test: Hyperbola- 2 - Question 6

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

Detailed Solution for Test: Hyperbola- 2 - Question 6

Given, 4x - 3y = 5 and 2x2 - 3y2 = 12

Test: Hyperbola- 2 - Question 7

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

Detailed Solution for Test: Hyperbola- 2 - Question 7

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.

Test: Hyperbola- 2 - Question 8

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

Test: Hyperbola- 2 - Question 9

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 =

Detailed Solution for Test: Hyperbola- 2 - Question 9

 is nearest to 3x + 4y -10 = 0

 
CP = ae = 10

Test: Hyperbola- 2 - Question 10

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

Detailed Solution for Test: Hyperbola- 2 - Question 10

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

Test: Hyperbola- 2 - Question 11

Consider a branch of the hyperbola x2 - 2y2 - 2y - 6 = 0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is 

Test: Hyperbola- 2 - Question 12

If α + β = 3π then the chord joining the points 'a' and 'b' for hyperbola passes through 

Detailed Solution for Test: Hyperbola- 2 - Question 12

equation of the chord joining a andb is

it passes through (0, 0)

Test: Hyperbola- 2 - Question 13

The number of tangents and normals to the hyperbola  of the slope 1 is

Detailed Solution for Test: Hyperbola- 2 - Question 13

Equation of tangents and normal with slope 'm' will be 

Test: Hyperbola- 2 - Question 14

From any point on the hyperbola  tangents are drawn to the hyperbola  The area cut off by the chord of contact on the asymptotes is equal to

Detailed Solution for Test: Hyperbola- 2 - Question 14

Let (x1 y1) be point on 
chord to 

Area formed by the lines = 4ab

Test: Hyperbola- 2 - Question 15

If the line ax + by + c = 0 is a normal to the hyperbola x y = 1, then

Test: Hyperbola- 2 - Question 16

A tangent to the parabola x2 = 4ay meets the hyperbola x2 - y2 = a2 at two points P and Q, then midpoint of P and Q lies on the curve

Detailed Solution for Test: Hyperbola- 2 - Question 16

Equation of tangent to parabola y = mx- am2 .......(1) equation of chord of hyperbola whose midpoint is (h, k) is  hx - ky = h2 - k2 ...... (2) form (1) and  (2) 

Test: Hyperbola- 2 - Question 17

The point of intersection of two tangents to the hyperbola x2/a2 – y2/b2 = 1, the product of whose slopes is c2, lies on the curve.

Test: Hyperbola- 2 - Question 18

Let P(a secθ, b tanθ) and Q(a secφ, b tanφ) where θ+φ = π/2, be two points on the hyperbola If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to

Detailed Solution for Test: Hyperbola- 2 - Question 18

Equation of normal at P (θ) is

ax+bycosecθ = (a2 +b2)secθ  .......(1)
And at Q (φ) is
ax + bycoseφ = (a2 +b2) secφ
⇒ ax +bysecθ = (a2 +b2)cosecθ   ......(2)
From (1) and (2) 

Test: Hyperbola- 2 - Question 19

Foot of normals drawn from the point p(h,k) to the hyperbola  will always lie on the conic

Detailed Solution for Test: Hyperbola- 2 - Question 19

Equation of normal at any point (x1, y1) is  it passes through 
P (h, k) then  thus (x1, y1) lie on the conic 

Test: Hyperbola- 2 - Question 20

A variable chord PQ, x cos θ + y sin θ = P of the hyperbola  subtends a right angle at the origin. This chord will always touch a circle whose radius is 

Detailed Solution for Test: Hyperbola- 2 - Question 20

Combined equation of lines joining the points of intersection of the chord with origin is 



these lines are mutually perpendicular if 

thus chord becomes x cos θ + y sin θ = which is clearly a tangent to the circle having centre at origin and radius is equal to

Test: Hyperbola- 2 - Question 21

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

Detailed Solution for Test: Hyperbola- 2 - Question 21

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 

Test: Hyperbola- 2 - Question 22

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

Detailed Solution for Test: Hyperbola- 2 - Question 22

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

Test: Hyperbola- 2 - Question 23

The asymptotes of the hyperbola hx + ky = xy are

Detailed Solution for Test: Hyperbola- 2 - Question 23

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

Test: Hyperbola- 2 - Question 24

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

Detailed Solution for Test: Hyperbola- 2 - Question 24

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

Test: Hyperbola- 2 - Question 25

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

Detailed Solution for Test: Hyperbola- 2 - Question 25

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

Test: Hyperbola- 2 - Question 26

If the curve  cut each other orthogonally then 

Detailed Solution for Test: Hyperbola- 2 - Question 26



Slope of 


 ------(1)
now solving the curves

-------(2)
from (1) & (2)

Test: Hyperbola- 2 - Question 27

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

Detailed Solution for Test: Hyperbola- 2 - Question 27

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

⇒ ax2 + ay2 + c = 0 represent a circle

Test: Hyperbola- 2 - Question 28

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

Detailed Solution for Test: Hyperbola- 2 - Question 28

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

Test: Hyperbola- 2 - Question 29

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

Detailed Solution for Test: Hyperbola- 2 - Question 29

Asymptotes are equally inclined to the axes of hyperbola Find the bisector of the asymptotes which bisects the angle containing the origin.

Test: Hyperbola- 2 - Question 30

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

Detailed Solution for Test: Hyperbola- 2 - Question 30

a = ae, i.e., 

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