Test: Hyperbola- 1 - JEE MCQ

# Test: Hyperbola- 1 - JEE MCQ

Test Description

## 20 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Hyperbola- 1

Test: Hyperbola- 1 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Hyperbola- 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Hyperbola- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hyperbola- 1 below.
Solutions of Test: Hyperbola- 1 questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Hyperbola- 1 solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Hyperbola- 1 | 20 questions in 40 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Hyperbola- 1 - Question 1

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

Detailed Solution for Test: Hyperbola- 1 - Question 1

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

Test: Hyperbola- 1 - Question 2

### 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

Detailed Solution for Test: Hyperbola- 1 - Question 2

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.

 1 Crore+ students have signed up on EduRev. Have you?
Test: Hyperbola- 1 - Question 3

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

Detailed Solution for Test: Hyperbola- 1 - Question 3

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
⇒ 4x2−5y=100

Test: Hyperbola- 1 - Question 4

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

Detailed Solution for Test: Hyperbola- 1 - Question 4

Test: Hyperbola- 1 - Question 5

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

Detailed Solution for Test: Hyperbola- 1 - Question 5

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

Test: Hyperbola- 1 - Question 6

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

Detailed Solution for Test: Hyperbola- 1 - Question 6

Test: Hyperbola- 1 - Question 7

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

Detailed Solution for Test: Hyperbola- 1 - Question 7

Test: Hyperbola- 1 - Question 8

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

Detailed Solution for Test: Hyperbola- 1 - Question 8

Test: Hyperbola- 1 - Question 9

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

Detailed Solution for Test: Hyperbola- 1 - Question 9

Test: Hyperbola- 1 - Question 10

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

Detailed Solution for Test: Hyperbola- 1 - Question 10

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

Test: Hyperbola- 1 - Question 11

The equation to the common tangents to the two hyperbolas  and  are

Detailed Solution for Test: Hyperbola- 1 - Question 11

Test: Hyperbola- 1 - Question 12

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

Detailed Solution for Test: Hyperbola- 1 - Question 12

Test: Hyperbola- 1 - Question 13

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

Detailed Solution for Test: Hyperbola- 1 - Question 13

Test: Hyperbola- 1 - Question 14

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

Detailed Solution for Test: Hyperbola- 1 - Question 14

Test: Hyperbola- 1 - Question 15

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

Detailed Solution for Test: Hyperbola- 1 - Question 15

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

Test: Hyperbola- 1 - Question 16

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

Detailed Solution for Test: Hyperbola- 1 - Question 16

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

Test: Hyperbola- 1 - Question 17

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

Detailed Solution for Test: Hyperbola- 1 - Question 17

e2 = (a/ 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

Test: Hyperbola- 1 - Question 18

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

Detailed Solution for Test: Hyperbola- 1 - Question 18

Test: Hyperbola- 1 - Question 19

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

Detailed Solution for Test: Hyperbola- 1 - Question 19

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

Test: Hyperbola- 1 - Question 20

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

Detailed Solution for Test: Hyperbola- 1 - Question 20

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

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests
Information about Test: Hyperbola- 1 Page
In this test you can find the Exam questions for Test: Hyperbola- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Hyperbola- 1 , EduRev gives you an ample number of Online tests for practice

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests