JEE Advanced Maths Hyperbola Previous Year Questions with Solutions

The JEE Advanced Maths Hyperbola previous year questions and solutions are given on this page. These are solved by our top JEE subject experts. These question papers given here will help candidates to understand the question types along with the paper pattern and the difficulty level of questions from each chapter. This helps them to self-analyse their preparation level.

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JEE Advanced Maths Hyperbola Previous Year Questions with Solutions

Question 1: If 2x – y + 1 = 0 is a tangent to the hyperbola x²/a² – y²/16 = 1, then which of the following cannot be sides of a right angles triangle?

(a) a, 4, 1

(b) a, 4, 2

(c) 2a, 8, 1

(d) 2a, 4, 1

Solution:

Given 2x – y + 1 = 0 is a tangent to the hyperbola x²/a² – y²/16 = 1.

The general form of tangent is y = mx + c where condition of tangency is

c² = a²m² – b² ..(i)

From the equation of hyperbola, we get a² = a²and b² = 16

From the equation of tangent, we get m = 2 and c = 1

Put the values in (i), we get

1² = 4a² – 16

=> 4a² = 17

So a = √17/2

Now we check the options which satisfy the Pythagoras theorem.

Check option (d) 2a, 4, 1.

=> √17, 4, 1 ; 4² + 1² = √17²

Option d satisfies the Pythagoras theorem.

Option a, b, c do not satisfy the Pythagoras theorem.

Hence option a, b, and c are correct.

Question 2: A hyperbola passes through the foci of the ellipse x²/25 + y²/16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

(a) x²/9 – y²/4 = 1

(b) x²/9 – y²/16 = 1

(c) x² – y² = 9

(d) x²/9 – y²/25 = 1

Solution:

Given equation of ellipse is x²/25 + y²/16 = 1

=> a = 5, b = 4

For ellipse, e₁ = √(1 – b²/a²)

= √(1-16/25)

=3/5

Foci = (±3,0)

Let equation of hyperbola be x²/A² – y²/B² = 1, passes through (±3, 0)

=> A² = 9

A = 3

e₂ = 5/3

e₂² = 1 + B²/A²

25/9 = 1 + B²/9

B² = 16

x²/9 – y²/16 = 1

Hence option b is the answer.

Question 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

Given (√3)kx + ky – 4√3 = 0

=> k (√3x + y) = 4√3

=> k = 4√3/(√3x + y) …(i)

Given √3x – y – 4√3 k = 0

=> k = (√3x – y)/4√3…(ii)

Equating (i) and (ii)

4√3/(√3x + y) = (√3x – y)/4√3

=> 48 = 3x² – y²

=> x²/16 – y²/48 = 1

This is the equation of a hyperbola with a² = 16 and b² = 48

Eccentricity = √(1 + b²/a²)

e² = 1 + 48/16

= 1 + 3

= 4

So e = 2

Thus the eccentricity is 2.

Hence option c is the answer.

Question 4: If the line y = mx + 7√3 is normal to the hyperbola x²/24 – y²/18 = 1, then a value of m is

(a) √5/2

(b) √15/2

(c) 2/√5

(d) 3/√5

Solution:

Since lx + my + n = 0 is a normal to x²/a² – y²/b² = 1

Then a²/l² – b²/m² = (a²+b²)²/n²

Given that line y = mx + 7√3 is normal to the hyperbola x²/24 – y²/18 = 1

24/m² – 18/(-1)² = (24 + 18)²/(7√3)²

=> m = 2/√5

Hence option c is the answer.

Question 5: If e₁ and e₂ are the eccentricities of the ellipse, (x²/18) + (y²/4) = 1 and the hyperbola, (x²/9) – (y²/4) = 1 respectively and (e₁, e₂) is a point on the ellipse, 15x² + 3y² = k. Then k is equal to:

(a) 14

(b) 15

(c) 17

(d) 16

Solution:

Given ellipse (x²/18) + (y²/4) = 1

Comparing with standard equation of ellipse,

We get a² = 18, b² = 4

Eccentricity, e = √(1 – b²/a²)

e₁ = √(1-(4/18))

= √7/3

Given hyperbola (x²/9) – (y²/4) = 1

Eccentricity, e = √(1 + b²/a²)

e₂ = √(1 + (4/9))

= √13/3

Since (e₁, e₂) lies on the ellipse 15x² + 3y² = k

15×(7/9) + 3×(13/9) = k

k = 16

Hence option d is the answer.

Question 6: Tangents are drawn to the hyperbola x²/9 – y²/4 = 1, parallel to the straight line 2x – y = 1. The points of contact of the tangents on the hyperbola are

(a) (9/2√2, 1/√2)

(b) (-9/2√2, -1/√2)

(c) (3√3, -2√2)

(d) (-3√3, 2√2)

Solution:

If slope of tangents to hyperbola x²/a² – y²/b² = 1 is m,

then equations of tangent to the hyperbola is y = mx ± √(a²m² – b²) with the points of contact is (±a²m/√(a²m² – b²), ±b²/√(a²m² – b²))

Tangent to hyperbola x²/9 – y²/4 = 1 is parallel to 2x – y = 1

Slope of tangent = 2

Point of contact is (±9×2/√(9×4 – 4), ±4/√(9×4 – 4))

= (9/2√2, 1/√2) and (-9/2√2, -1/√2)

Hence option a and b are correct.

Question 7: Let the eccentricity of the hyperbola x²/a² – y²/b² = 1 be the reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then

(a) the equation of the hyperbola is x²/3 – y²/2 = 1

(b) a focus of the hyperbola is (2, 0)

(c) the eccentricity of the hyperbola is √(5/3)

(d) the equation of the hyperbola is x² – 3y² = 3

Solution:

The given equation of the ellipse is x² + 4y² = 4

(x²/4) + (y²/1) = 1.

The eccentricity of the ellipse, e = √(1 – (b²/a²))

= √(1 – (1/4))

= √3/2

The eccentricity of the hyperbola = 2/√3 (reciprocal)

√(1 + (b²/a²) ) = 2/√3

1 + (b²/a²) = 4/3

(b/a) = 1/√3 ..(i)

Focus of the ellipse = (± ae, 0)

= (± √3, 0)

The hyperbola passes through (± √3, 0).

3/a² = 1

a = √3

From equation (1), b = 1.

The equation of the hyperbola is (x²/3) – (y²/1) = 1.

x² – 3y² = 3

Focus of the hyperbola = (± ae, 0)

= (± 2, 0).

Hence option b and d are correct.

Question 8: Let a hyperbola passes through the focus of the ellipse x²/25 + y²/16 = 1. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

(a) the equation of hyperbola is x²/9 – y²/16 = 1

(b) the equation of hyperbola is x²/9 – y²/25 = 1

(c) focus of hyperbola (5, 0)

(d) vertex of hyperbola is (5√3, 0)

Solution:

For the given ellipse x²/25 + y²/16 = 1

=> e = √(1 – 16/25)

= 3/5

Eccentricity of hyperbola = 5/3

Let the hyperbola be x²/A² – y²/B² = 1, then

B² = A²(25/9 – 1)

= (16/9)A²

So (x²/A²) – (9y²/16A²) = 1

As it passes through focus of ellipse (3, 0)

We get A² = 9

=> B² = 16

Equation of hyperbola is (x²/9) – (y²/16) = 1

Its focus is (5, 0) and vertex is (3, 0).

Hence option a and c are correct.

Question 9: The line 2x + y = 1 is a tangent to the hyperbola x²/a² – y²/b² = 1. If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

(a) 2

(b) -1

(c) 4

(d) √2

Solution:

Given hyperbola x²/a² – y²/b² = 1

Tangent is 2x + y = 1

Here m = -2, c = 1

Intersection point of nearest directrix x = a/e and x axis is (a/e, 0).

Since 2x+y = 1 passes through (a/e, 0)

2a/e = 1

=> a = e/2

For tangency, c² = a²m² – b²

1 = a²(-2)² – b²

4a² – b² = 1

4a² – a²(e²-1) = 1

Since a = e/2, above equation becomes

e² – (e²/4)(e² – 1) = 1

e⁴ – 5e² + 4 = 0

(e² – 4)(e² – 1) = 0

=> e² = 4 or e² = 1

Since e>1 for hyperbola, e = 2

Hence option a is the answer.

Question 10: The equation x²/(1-r) – y²/(1+r) = 1, r>1 represents the

(a) ellipse

(b) hyperbola

(c) circle

(d) none of these

Solution:

Given that x²/(1-r) – y²/(1+r) = 1, r>1

Since r>1, 1-r < 0 and 1+r>0

Let 1-r = -a², 1+r = b²

So (x²/-a²) – (y²/b²) = 1

=> (x²/a²) + (y²/b²) = -1, which is not possible for any real values of x and y.

Hence option d is the answer.

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