Examples Based on Hyperbola

 

Illustration 1: If the chords of the hyperbola x² – y² = a² touch the parabola y² = 4ax, then the locus of the middle points of these chords is the curve

(a) y² (x + a) = x³ 

(b) y² (x – a) = x³ 

(c) y² (x + 2a) = 3x³ 

(d) y² (x – 2a) = 2x³ 

Solution: Let the mid-point of the chord be (h,k)

Then, the equation of the chord of x² – y²  = a² is

T = S₁

hx – ky = h² - k²

So, y = h/k x – (h² - k²)/k and this is tangent of y² = 4ax

so, – (h² – k²)/k = a/(h/k)

or -(h² - k²) = ak²/h

or -h³ + hk² = ak²

or k²(h-a) = h³

Hence, locus of mid-point is y²(x-a) = x³.

 

Illustration 2: Find the equation of the hyperbola whose directrix is 2x + y = 1, focus is (1, 1) and eccentricity is √3.

Solution: Let S(1, 1) be focus and P(x, y) be any point on the hyperbola. From P draw PM perpendicular to the directrix then PM = (2x + y – 1)/√(2² + 1²) = (2x + y – 1)/√5

Also from the definition of the hyperbola, we have

SP/PM = e ⇒ SP = ePM

⇒ √(x–1)² + (y–1)² = √3 (2x + y – 1/√5)

⇒ (x – 1)² + (y – 1)² = 3 (2x + y – 1)2/5

⇒ 5[(x² + 1 – 2x) + (y² + 1 – 2y)] = 3(4x² + y² + 1 + 4xy – 4x – 2y)

⇒ 7x² – 2y² + 12xy – 2x – 4y – 7 = 0

 

Illustration 3: The normal at P to a hyperbola of eccentricity e intersects its transverse and conjugate axes at L and M respectively. If locus of the mid-point of LM is hyperbola, then eccentricity of the hyperbola is 

(a) (e + 1)/(e - 1)

(b) e/√(e² - 1)

(c) e

(d) none of these

Solution: Equation of normal at P (a sec φ, b tan φ) is

ax cos φ + by cot φ = a² + b²

Hence, the coordinates of L and M are

L = [(a² + b²)sec φ/a, 0 ]

and M = [0, (a² + b²)tan φ/b] respectively

Let mid point of LM be Q(h,k)

then h = (a² + b²)sec φ/2a

So, sec φ = 2ah/(a²+b²)

and k = (a²+b²)tan φ/2b

so, tan φ = 2bk/(a²+b²)

so, we have sec² φ – tan²φ = 4a²h²/(a²+b²)² – 4b²k²/(a²+b²)²

Hence, the locus

x²/[(a²+b²)/2a]² – y²/[(a²+b²)/2b]² = 1

Suppose eccentricity is e1.

then, [(a²+b²)/2b]² = [(a²+b²)/2a]² (e₁² – 1)

So, a² = b² (e₁² -1)

a² = a²(e²-1)(e₁²-1)

So, e²e₁² – e²  -e₁² + 1 = 1

So, e₁² (e²-1) = e²

e₁ = e/√(e²-1)

 

Illustration 4: The equations to the common tangents to the two hyperbolas x²/a² – y²/b² = 1 and y²/a² – x²/b² = 1 are

(a) y = ± x ± √(b² - a²)

(b) y = ± x ± √(a² - b²)

(c) y = ± x ± (a² – b²)

(d) y = ± x ± √(b² + a²)

Solution: Let y = mx+c be a common tangent of hyperbola

x²/a² – y²/b² = 1 and

y²/a² – x²/b² = 1

The condition of tangency for the first equation is c² = a²m² – b²     ….. (1)

and the condition of tangency for the second equation is c² = a² - b²m²

From these two equations it follows that,

a²m² – b² = a² – b²m²

So, a²(m² – 1) + b²(m² – 1) = 0

so, (a² + b²)(m² - 1) = 0

since, a² + b² ≠ 0, so m² - 1 = 0

So, m = ± 1.

From eq.(1) c² = a² - b²

So, c = ± √(a² – b²)

Hence, the equations of common tangents are y = ± x ± √(a² – b²)

 

Illustration 5: Find the directrix, foci and eccentricity of the hyperbola ax² – y² = 1

Solution: The given hyperbola is ax² – y² = 1

or     x²/(1/a) – y²/1 = 1                                                         …… (1)

which is of the form (x²/a²) – (y²/b²) = 1

Here a² = 1/a, b² = 1

If e be the eccentricity of the hyperbola, then

b² = a²(e² – 1)

⇒ 1 =  (e² – 1)

⇒ a = (e² – 1)

or e² = a + 1 or e = √(a + 1)

Also the foci are given by (+ ae, 0)

∴ The required foci are (+ 1/√a √(a+1), 0)

or (+ √a+1/a, 0)

And the directrix is given by x = + (a/e)

⇒ x = + [1/√a/√a+1]                    (since a = 1/√a, e = 1/√a+1)

x = + 1/√a(a+1)

Illustration 6: Find the locus of a point, the difference of whose distances from two fixed points is constant.

Solution: Let two fixed points be S (ae, 0) and S’ (–ae, 0).

Let P(x, y) be a moving point such that

SP – S’P = Constant = 2a (say).

Then √[(x – ae)² + (y – 0)²] – √[x + ae]² + (y – 0)² = 2a

⇒ √[(x – ae)² + y²] = + 2a + √[(x + ae)² + y²

⇒ (x – ae)² + y² = 4a² + (x + a²)² + y² + 4a √[(x – ae)² + y²]

⇒ (x – ae)² – (x + ae)² – 4a² = + 4a √[(x – ae)² + y²]

⇒ –4aex – 4a² = + 4a √[(x – ae)² + y²]

⇒ (ex + a)² = (x + ae)² + y²

⇒ (e² – 1)x² – y² = a²(e² – 1)

⇒ x²/a² – y²/b² = 1 taking b² = a²(e² – 1)

This is a hyperbola.

 

Illustration 7: If A, B, C are three points on the hyperbola xy = c² and AC is perpendicular to BC, prove that AB is parallel to the normal to the curve at C.

Solution: Let the three points A, B, C respectively be (ct₁, c/t₁), (ct₂, c/t₂) and (ct₃, c/t₃). Since AC is perpendicular to BC,

(c/t₃–c/t₁/ct₃–ct₁) = – 1 ⇒ t₁t₂ = –1                                 …… (1)

Normal to the curve at C (ct₃, c/t₃) is

y = xt₃² + 2/t₃ (1 – t₃⁴) and its slope is t₃² = –1/t₁t₂              …… (2)

Slope of AB = (c/t₂–c/t₁/ct₂–ct₁) = – 1/t₁t₂ = t₃²

⇒ AB is parallel to the normal at C.

Illustration 8: Find the equation of the hyperbola the distance between whose foci is 16, whose eccentricity is √2 and whose axis is along the x-axis centre being the origin.

Solution: We have b² = a²(e² – 1) = a² 

⇒ b = a.

Also 2ae = 16 ⇒ ae = 8 ⇒ a = 4√2.

Hence the equation of the required hyperbola is

x²/32 – y²/32 = 1 ⇒ x² – y² = 32.

Illustration 9: The perpendiculars drawn from the centre of a hyperbola x²/a² – y²/b² = 1 upon the tangent and normal at any point of the hyperbola meet them in Q and R. Find the locus of Q and R.

Solution: Tangent at any point P(a sec θ, b tan θ) is  sec θ – y/b  tan θ = 1. … (1)

Equation of the line through centre (origin) perpendicular to (1) is y = –a sin θ/b x

⇒ sin θ = – by/ax

Eliminating θ from (1), we get x/a cos θ – y/b cos θ (–by/ax) = 1.

⇒ x² + y² = ax cos θ ⇒ (x² + y²)² = a²x²(1 – b²y²/a²x²)

Or (x² + y²)² = a²x² – b²y², which is the locus of Q.

Normal at the point P (a sec θ, b tan θ) is ax cos θ + by

cot θ = a² + b²                                                         … (2)

Equation of the line perpendicular to (2) drawn from the centre is

y = bx/a sin θ                                                     … (3)

Form (2) and (3),

sin θ = bx/ay and ax √1–b²y²/a²x² + by √1–b²x²/a²y² . ay/bx = a² + b²

⇒ (x² + y²)² (a²y² – b²x²) = (a² + b²)² x²y², which is the locus of R.

 

Illustration 10: A straight line touches the rectangular hyperbola 9x² - 9y²  = 8 and the parabola y² = 32x. The equation of the line is:

(a) 9x + 3y – 8 = 0

(b) 9x - 3y + 8 = 0

(c) 9x + 3y + 8 = 0

(d) 9x - 3y – 8 = 0

Solution: Equation of tangent in terms of slope of y² = 32x is

y = mx + 8/m  …... (1)

which is also a tangent of 9x² – 9y² = 8

So, x² – y² = 8/9

Hence, (8/m)² = 8/9 m² – 8/9

So, 8/m² = m²/9 – 1/9

so, 72 = m⁴ – m²

So, m⁴ – m² – 72 = 0

(m² – 9)(m² + 8) = 0

So, m = 9 and m² + 8 ≠ 0

So, m = ± 3

So, from eq (1), we have y = ± 3x ± 8/3

So, 3y = ± 9x ± 8

or ± 9x – 3y ± 8 = 0

i.e. 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

 -9x – 3y + 8 = 0, – 9x – 3y – 8 = 0

or 9x – 3y + 8 = 0, 9x – 3y – 8 = 0

9x + 3y - 8 = 0, 9x + 3y + 8 = 0

 

Illustration 11: Assertion/Reason Based:

Choose the correct option out of the following four:

(a) Both A and R are individually true and R is the correct explanation of A.

(b) Both A and R are individually true but R is not the correct explanation of A.

(c) A is true but R is false

(d) A is false but R is true.

Assertion: Director circle of hyperbola x²/a² – y²/b² + 1 = 0 is defined only when b > a.

Reason: Director circle of hyperbola x²/25 – y²/9 = 1 is x² + y² = 16.

(a) A

(b) B

(c) C

(d) D

Solution: Since, the director circle is the locus of point of intersection of perpendicular tangents

Any tangent in terms of m of x²/a² – y²/b² + 1 = 0 is

y = mx ± √(b² - a²m²)

or (y – mx)² = b² – a²m²

so, m²(x² + a²) -2mxy + y² - b² = 0

since, m₁m₂ = -1

So, (y² - b²)/(x² + a²) = -1

or x² + y² = b² - a², b > a

Also, director circle of x²/25 – y²/9 = 1 is

x² + y² = 25 – 9 = 16.

 

Illustration 12: C is a centre of the hyperbola x²/a² – y²/b² = 1 and the tangent at any point P meets asymptotes in the point Q and R. Find the equation to locus of the centre of the circle circumscribing the triangle CQR.

Solution: This equation to the hyperbola is given as

x²/a² – y²/b² = 1                                      …… (1)

Let P any point on it as (a sec θ, b tan θ), then the equation of tangent at P is

x/a – y/b sin θ = cos θ                              …… (2)

The equation to the asymptotes to (1) are

x/a = y/b                                                …… (3)

and x/a = – y/b                                             …… (4)

Solving (2) and (3), we get the coordinates of Q as

(a cos θ/1–sin θ, bcosθ/1–sinθ)

Solving (2) and (4), we get the coordinates of R as

(acosθ/1+sinθ, –bcosθ/1+sinθ)

Let O be the centre of the circle passing through C, Q and R having its coordinates as (h, k). Then clearly OC = OQ

⇒ h² + k² = (h–acosθ/1–sinθ)² + (k–bcosθ/1–sinθ)²

⇒ h² + k² = (h–acosθ/1–sinθ)² + (k–bcosθ/1–sinθ)²

⇒ h² + k² = h² + k² + (a² + b²) cos²θ/(1–sinθ)² – (2ah + 2bk) cosθ/1–sinθ

⇒ 2(ah + bk) = (a² + b²) cosθ/1–sinθ                              …… (5)

Similarly OC = OR

Hence h² + k² = (h–acosθ/1+sinθ)² + (k+bcosθ/1+sinθ)²

Which on simplification as in the last case, given 2(ah –bk) = (a² + b²) cosθ/1–sinθ                                   …… (6)

to get the locus of the point O we have to eliminate f from (5) and (6), so multiplying the two we get

4(a²h² – b²k²) = (a² + b²) cos²θ/1–sin²θ = (a² + b²)

for (h, k), we get the required locus as

4(a²x² – b²y²) = (a² + b²)²

 

Illustration 13: The equation of a hyperbola conjugate to the hyperbola x² + 3xy + 2y² + 2x + 3y = 0 is

(a) x² + 3xy + 2y² + 2x + 3y + 1 = 0

(b) x² + 3xy + 2y² + 2x + 3y + 2 = 0

(c) x² + 3xy + 2y² + 2x + 3y + 3 = 0

(d) x² + 3xy + 2y² + 2x + 3y + 4 = 0

Solution: Since, the given hyperbola is H : x² + 3xy + 2y² + 2x + 3y = 0

Let the pair of asymptotes be x² + 3xy + 2y² + 2x + 3y + λ = 0

So, Δ = 0

1.2.λ + 2.3/2.1.3/2 – 1.9/4 – 2.1 - λ.9/4 = 0

-λ/4 + 9/2 - 9/4 – 2 = 0

λ/4 = 9/4 – 2 = ¼

So, λ = 1

Therefore, A : x² + 3xy + 2y² + 2x + 3y + 1 = 0

since, H + C = 2A

So, C = 2A – H

         = x² + 3xy + 2y² + 2x + 3y + 2

Hence, the conjugate hyperbola is x² + 3xy + 2y² + 2x + 3y + 2 = 0

 

Illustration 14: The locus of the point of intersection of two perpendicular tangents to the hyperbola x²/a² – y²/b² = 1 is

(a) director circle

(b) x² + y² = a²

(c) x² + y² = a² - b²

(d) x² + y² = a² + b²

Solution: Equation of any tangent in terms of slope m is

y = mx + (a²m² – b²)

It passes through (h, k), so we have

(k - mh)² = a²m² – b²

So, m²(h² - a²) – 2mhk + k² + b² = 0

This is a quadratic in m

Let the slopes of tangents be m₁ and m₂.

then m₁m₂ = -1.

So, (k² + b²)/(h² – a²) = -1

(h² + k²) = (a² – b²)

Hence, the locus is (x² + y²) = (a² – b²) which is the director circle of x²/a² – y²/b² = 1.

 

Illustration 15: If a tangent to the hyperbola x²/25 – y²/16 = 1 makes equal intercepts of lengths λ on the axes, then what is the value of λ?

Solution: Equation of a tangent at (5 sec θ, 4 tan θ) to the given hyperbola is x/5 sec θ – y/4 tan θ = 1

This intersects the axes at (5 cos θ, 0) and (0, – 4 cot θ).

So, we have  λ = |5 cos θ| = |4 cot θ|

So, λ² = 25 cos²θ = 16 cot²θ

So, sec²θ = 25/ λ² and tan²θ = 16/λ²

So, sec²θ – tan²θ = 25/λ² – 16/λ² =1

so, λ² = 25 - 16 = 9

So, λ = 3.

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