**JEE Previous Year Questions With Solutions on Hyperbola** are provided here in PDF format, which can be downloaded for free. In order to help students, understand the concepts clearly, we are offering a comprehensive set of previous year questions with solutions. These solutions will also help students boost their preparation for JEE Main and Advanced. Students will further get to learn concepts like: equation of hyperbola under different conditions, normal and tangents of hyperbola, Vertex, focii, eccentricity, axes, applications of hyperbola and many more. **Previous year chapter-wise solutions** will help students understand the concepts in a better way and pattern of questions. Students can download the pdf by clicking on the given link below and start practice.

Download Hyperbola Previous Year Solved Questions PDF

**JEE Main Past Year Questions With Solutions on Hyperbola**

**Question 1:** The locus of a point P(α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 is

(a) an ellipse

(b) a circle

(c) a hyperbola

(d) a parabola

**Answer: (c)**

**Solution: **

Tangent to the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 is y = mx ± √(a^{2}m^{2} – b^{2})

Given that y = αx + β is the tangent of hyperbola.

m = α and a^{2}m^{2} – b^{2} = β^{2}

Therefore, a^{2} α^{ 2} – b^{2} = β^{2}

Locus is a^{2} x^{ 2} – y^{2} = b^{2 }, which is parabola.

**Question 2:** If a hyperbola passes through the point P(10, 16) and it has vertices at (±6, 0), then the equation of the normal at P is:

(a) 3x + 4y = 94

(b) x + 2𝑦 = 42

(c) 2x + 5y = 100

(d) x + 3y = 58

**Answer: (c)**

**Solution:**

Vertex of hyperbola: (±a, 0) = (± 6, 0) => a = 6

We know, equation of hyperbola, x^{2}/a^{2} – y^{2}/b^{2} = 1

=> x^{2}/36 – y^{2}/b^{2} = 1

Point P(10,16) lies on parabola, so

100/36 – 256/b^{2} = 1

=> 64/36 = 256/b^{2}

=>b^{2} = 144

Equation of hyperbola becomes, x^{2}/36 – y^{2}/144 = 1 and

Equation of normal : a^{2}x/x_{1} + b^{2}y/y_{1} = a^{2} + b^{2}

=> 36x/10 + 144y/16 = 180

=> x/50 + y/20 = 1

Or 2x + 5y = 100

**Question 3: **The eccentricity of the hyperbola whose latus rectum is 8 and length of the conjugate axis is equal to half the distance between the foci, is

(a) √3 (b) 4/3 (c) 2/√3 (d) 4/√3

**Answer: (c)**

**Solution: **

We know, conjugate axis of hyperbola = 2b and

Latus rectum = 2b^{2}/a

Given: The eccentricity of the hyperbola whose latus rectum is 8 and length of the conjugate axis is equal to half the distance between the foci.

=> 2b = 1/2(2ae) and 2b^{2}/a = 8

=>2/a(ae/2)^{2 }= 8

=> ae^{2} = 16 …(i)

Also, we know, b^{2} = a^{2} (e^{2} – 1)

From equation, 2b^{2}/a = 8 =>b^{2} = 4a

So, a^{2} (e^{2} – 1) = 4a

=>ae^{2} – a = 4

Using (i)

=>16 – a = 4

Or a = 12

Again, (i)=> 12 e^{2} = 16

=> e = 2/√3

**Question 4: **An ellipse passes through the foci of the hyperbola, 9x^{2} −4y^{2} =36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 1/2. Find the equation of ellipse.

**Solution: **

Equation of hyperbola is 9x^{2} −4y^{2} =36 or x^{2}/4 − y^{2}/9 = 1

(Here a < b)

Focus = (0, ± be)

Eccentricity = e = √(1+4/9 = √13/3

So, Foci of hyperbola: (0, ±√13)

Standard equation of the ellipse, x^{2}/a^{2} + y^{2}/b^{2} = 1 …(i)

Eccentricity = e’ = √(1-a^{2}/b^{2}) …(ii)

ee’ = 1/2 (given)

Using eccentricity value of hyperbola, e’ = 1/2 x 3/√13 = 3/2√13

(ii) => e’^{2} = (1-a^{2}/b^{2})

9/52 = (1-a^{2}/b^{2})

Find the value of b2 form (i) using focii 13/b^{2} = 1 => b^{2} = 13

=> 9/52 = (1-a^{2}/13)

=> 9/4 = 13 – a^{2}

=> a^{2} = 43/4

Now equation of ellipse is 4x^{2}/43 + y^{2}/13 = 1

**Question 5:** If e_{1} and e_{2} are the eccentricities of the ellipse, x^{2}/18 + y^{2}/4 = 1 and the hyperbola, x^{2}/9 – y^{2}/4 = 1 respectively and (e_{1}, e_{2}) is a point on the ellipse, 15x^{2} + 3y^{2} = k. Then k is equal to

(a) 14 (b) 15 (c) 17 (d) 16

**Answer: (d)**

**Solution: **

e_{1} and e_{2} are the eccentricities of the ellipse, x^{2}/18 + y^{2}/4 = 1 and the hyperbola, x^{2}/9 – y^{2}/4 = 1 respectively

As (e_{1} , e_{2}) lies on the ellipse 15x^{2} + 3y^{2} = k

Because, 15e_{1}^{2} + 3e_{2}^{2} = k

=> 15 x 7/9 + 3 x 13/9 = k

=> k = 16

**Question 6:** A tangent to the hyperbola x^{2}/4 – y^{2}/2 = 1 meets x -axis at P and y -axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). Then R lies on:

(a) 2/x^{2} – 4/y^{2} = 1

(b) 4/x^{2} – 2/y^{2} = 1

(c) 4/x^{2} + 2/y^{2} = 1

(d) 2/x^{2} + 4/y^{2} = 1

**Answer: (b)**

**Solution: **

Equation of tangent at point (a sec θ, b tanθ) is x/a secθ – y/b tanθ = 1

Equation of tangent of hyperbola x^{2}/4 – y^{2}/2 = 1 is (x secθ)/2 – (y tanθ)/√2 = 1 at any parametric point Q.

The coordinate of P and Q are (a cosθ, 0) and (0, -b cotθ) respectively.

OPRQ is a rectangle, Let the coordinates of point R be (h, k).

then h = a cos θ and k = -b cotθ

Therefore, secθ = a/h and tanθ = -b/k

=> a^{2}/h^{2} – b^{2}/k^{2} = 1

So, the required locus is 4/x^{2} – 2/y^{2} = 1

**Question 7:** If the eccentricity of the hyperbola

(a) (3, ∞) (b) (1, 3/2) (c) (2, 3) (d) (-3, -2)

**Answer: (a)**

**Solution: **

We know, for hyperbola, e^{2} = 1 + b^{2}/a^{2}

= 1 + tan^{2} θ = sec^{2} θ

If e > 2 => sec θ > 2

=> θ Є (π/3, π/2)

Now, length of latus rectum of hyperbola = 2b^{2}/a = 2 tan θ sin θ

= 2 > √3 > √3/2 > 3

Therefore, the values of length of latus rectum lies in the interval (3, ∞).

**Question 8:** Let the eccentricity of the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 be the reciprocal to that of the ellipse x^{2} + 4y^{2} = 4. If the hyperbola passes through a focus of the ellipse, then

(a) the equation of the hyperbola is x^{2}/3 – y^{2}/2 = 1

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

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

(d) the equation of the hyperbola is x^{2} – 3y^{2} = 3

**Answer: (d)**

**Solution: **

The eccentricity of ellipse, e = √(1-b^{2}/a^{2})

Given equation of ellipse x^{2} + 4y^{2} = 4 can be rewritten as x^{2}/4 + y^{2}/1 = 1.

eccentricity = √(1-1/4) = √3/2

Given: The eccentricity of the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 be the reciprocal to that of the ellipse x^{2} + 4y^{2} = 4.

=> eccentricity of hyperbola = 2/√3

Now,

=> √(1+b^{2}/a^{2}) = 2/√3

=> (1+b^{2}/a^{2}) = (2/√3)^{2}

=> b/a = 1/√3

Focus of ellipse = (±ae, 0) = (±√3, 0)

Hyperbola passes through Focus, so 3/a^{2} = 1 => a = √3

And b/a = 1/√3 = > b/√3 = 1/√3 => b = 1

The equation of hyperbola is x^{2}/3 – y^{2}/1 = 1

Or x^{2} – 3y^{2} = 3

Focus of hyperbola = (±ae, 0) = (±2, 0)

**Question 9:** Tangents are drawn to the hyperbola 4x^{2} – y^{2} = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then find the area of △PTQ. [Area in sq. units]

**Solution: **

Equation of hyperbola can be written as, 4x^{2}/36 – y^{2}/36 = 1

or x^{2}/9 – y^{2}/36 = 1

Let the equation of tangent at any point P(x_{1}, y_{1}) is xx_{1}/a^{2} – yy_{1}/b^{2} = 1

As tangent is passing through the point (0, 3) (Given)

So, 0/a^{2} – 3y/36 = 1

=> y = -12 and

Again, 4x^{2} – y^{2} = 36

Using value of y, we have x = ±3√5

So, the coordinates of P and Q are (3√5, -12) and (-3√5, -12) respectively.

Now, area od triangle TPQ,

= 1/2|3√5(-12 – 3) + 12 (-3√5 – 0) + 1(-9√5 – 0)|

= 1/2|-45√5 – 36√5 – 9√5|

= 1/2|90√5|

= 45√5

**Question 10:** The circle x^{2} + y^{2} = 8x and hyperbola x^{2}/9 – y^{2}/4 = 1 intersect at the points A and B. Find the equation of a common tangent with positive slope to the circle as well as to the hyperbola.

Solution:

The equation of circle x^{2} + y^{2} = 8x can be rewritten as (x – 4)^{2} + y^{2} = 16

Tangent to hyperbola is y = mx + √(9m^{2}-4), m> 0.

Distance from center to the tangent is:

on solving above equation, we get

m = 2/√5

=> y = 2x/√5 + 4/√5

Or 2x – √5 y + 4 = 0

Therefore, 2x – √5 y + 4 = 0 is equation of a common tangent with positive slope to the circle as well as to the hyperbola.

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