**Examples Based on Hyperbola**

**Illustration 1:** If the chords of the hyperbola x^{2} – y^{2} = a^{2} touch the parabola y^{2} = 4ax, then the locus of the middle points of these chords is the curve

(a) y^{2} (x + a) = x^{3}

(b) y^{2} (x – a) = x^{3}

(c) y^{2} (x + 2a) = 3x^{3}

(d) y^{2} (x – 2a) = 2x^{3}

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

Then, the equation of the chord of x^{2} – y^{2} = a^{2} is

T = S_{1}

hx – ky = h^{2 }- k^{2}

So, y = h/k x – (h^{2 }- k^{2})/k and this is tangent of y^{2} = 4ax

so, – (h^{2} – k^{2})/k = a/(h/k)

or -(h^{2} - k^{2}) = ak^{2}/h

or -h^{3} + hk^{2} = ak^{2}

or k^{2}(h-a) = h^{3}

Hence, locus of mid-point is y^{2}(x-a) = x^{3}.

**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)/√(2^{2} + 1^{2}) = (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[(x^{2} + 1 – 2x) + (y^{2} + 1 – 2y)] = 3(4x^{2} + y^{2} + 1 + 4xy – 4x – 2y)

⇒ 7x^{2} – 2y^{2} + 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/√(e^{2} - 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 = a^{2} + b^{2}

Hence, the coordinates of L and M are

L = [(a^{2} + b^{2})sec φ/a, 0 ]

and M = [0, (a^{2} + b^{2})tan φ/b] respectively

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

then h = (a^{2} + b^{2})sec φ/2a

So, sec φ = 2ah/(a^{2}+b^{2})

and k = (a^{2}+b^{2})tan φ/2b

so, tan φ = 2bk/(a^{2}+b^{2})

so, we have sec^{2} φ – tan^{2}φ = 4a^{2}h^{2}/(a^{2}+b^{2})^{2} – 4b^{2}k^{2}/(a^{2}+b^{2})^{2}

Hence, the locus

x^{2}/[(a^{2}+b^{2})/2a]^{2} – y^{2}/[(a^{2}+b^{2})/2b]^{2} = 1

Suppose eccentricity is e1.

then, [(a^{2}+b^{2})/2b]^{2} = [(a^{2}+b^{2})/2a]^{2} (e_{1}^{2} – 1)

So, a^{2} = b^{2} (e_{1}^{2} -1)

a^{2} = a^{2}(e^{2}-1)(e_{1}^{2}-1)

So, e^{2}e_{1}^{2} – e^{2} -e_{1}^{2} + 1 = 1

So, e_{1}^{2} (e^{2}-1) = e^{2}

e_{1} = e/√(e^{2}-1)

**Illustration 4:** The equations to the common tangents to the two hyperbolas x^{2}/a^{2} – y^{2}/b^{2} = 1 and y^{2}/a^{2} – x^{2}/b^{2} = 1 are

(a) y = ± x ± √(b^{2 }- a^{2})

(b) y = ± x ± √(a^{2 }- b^{2})

(c) y = ± x ± (a^{2} – b^{2})

(d) y = ± x ± √(b^{2} + a^{2})

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

x^{2}/a^{2} – y^{2}/b^{2} = 1 and

y^{2}/a^{2} – x^{2}/b^{2} = 1

The condition of tangency for the first equation is c^{2 }= a^{2}m^{2} – b^{2} ….. (1)

and the condition of tangency for the second equation is c^{2} = a^{2 }- b^{2}m^{2}

From these two equations it follows that,

a^{2}m^{2} – b^{2} = a^{2} – b^{2}m^{2}

So, a^{2}(m^{2} – 1) + b^{2}(m^{2} – 1) = 0

so, (a^{2} + b^{2})(m^{2 }- 1) = 0

since, a^{2 }+ b^{2} ≠ 0, so m^{2 }- 1 = 0

So, m = ± 1.

From eq.(1) c^{2} = a^{2 }- b^{2}

So, c = ± √(a^{2} – b^{2})

Hence, the equations of common tangents are y = ± x ± √(a^{2} – b^{2})

**Illustration 5:** Find the directrix, foci and eccentricity of the hyperbola ax^{2} – y^{2} = 1

**Solution:** The given hyperbola is ax^{2} – y^{2} = 1

or x^{2}/(1/a) – y^{2}/1 = 1 …… (1)

which is of the form (x^{2}/a^{2}) – (y^{2}/b^{2}) = 1

Here a^{2} = 1/a, b^{2} = 1

If e be the eccentricity of the hyperbola, then

b^{2} = a^{2}(e^{2} – 1)

⇒ 1 = (e^{2} – 1)

⇒ a = (e^{2} – 1)

or e^{2} = 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} + y^{2}] = + 2a + √[(x + ae)^{2} + y^{2}

⇒ (x – ae)^{2} + y^{2} = 4a^{2} + (x + a^{2})^{2} + y^{2} + 4a √[(x – ae)^{2} + y^{2}]

⇒ (x – ae)^{2} – (x + ae)^{2} – 4a^{2} = + 4a √[(x – ae)^{2} + y^{2}]

⇒ –4aex – 4a^{2} = + 4a √[(x – ae)^{2 }+ y^{2}]

⇒ (ex + a)^{2} = (x + ae)^{2} + y^{2}

⇒ (e^{2} – 1)x^{2} – y^{2} = a^{2}(e^{2} – 1)

⇒ x^{2}/a^{2 }– y^{2}/b^{2} = 1 taking b^{2} = a^{2}(e^{2} – 1)

This is a hyperbola.

**Illustration 7:** If A, B, C are three points on the hyperbola xy = c^{2} 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 (ct_{1}, c/t_{1}), (ct_{2}, c/t_{2}) and (ct_{3}, c/t_{3}). Since AC is perpendicular to BC,

(c/t_{3}–c/t_{1}/ct_{3}–ct_{1}) = – 1 ⇒ t_{1}t_{2} = –1 …… (1)

Normal to the curve at C (ct_{3}, c/t_{3}) is

y = xt_{3}^{2} + 2/t_{3} (1 – t_{3}^{4}) and its slope is t_{3}^{2} = –1/t_{1}t_{2} …… (2)

Slope of AB = (c/t_{2}–c/t_{1}/ct_{2}–ct_{1}) = – 1/t_{1}t_{2} = t_{3}^{2}

⇒ 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 b^{2} = a^{2}(e^{2} – 1) = a^{2}

⇒ b = a.

Also 2ae = 16 ⇒ ae = 8 ⇒ a = 4√2.

Hence the equation of the required hyperbola is

x^{2}/32 – y^{2}/32 = 1 ⇒ x^{2} – y^{2} = 32.

**Illustration 9:** The perpendiculars drawn from the centre of a hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 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.

⇒ x^{2} + y^{2} = ax cos θ ⇒ (x^{2} + y^{2})^{2} = a^{2}x^{2}(1 – b^{2}y^{2}/a^{2}x^{2})

Or (x^{2} + y^{2})^{2 }= a^{2}x^{2} – b^{2}y^{2}, which is the locus of Q.

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

cot θ = a^{2} + b^{2} … (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–b^{2}y^{2}/a^{2}x^{2} + by √1–b^{2}x^{2}/a^{2}y^{2} . ay/bx = a^{2} + b^{2}

⇒ (x^{2} + y^{2})^{2} (a^{2}y^{2} – b^{2}x^{2}) = (a^{2} + b^{2})^{2} x^{2}y^{2}, which is the locus of R.

**Illustration 10:** A straight line touches the rectangular hyperbola 9x^{2} - 9y^{2} = 8 and the parabola y^{2} = 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 y^{2} = 32x is

y = mx + 8/m …... (1)

which is also a tangent of 9x^{2} – 9y^{2} = 8

So, x^{2} – y^{2} = 8/9

Hence, (8/m)^{2} = 8/9 m^{2} – 8/9

So, 8/m^{2} = m^{2}/9 – 1/9

so, 72 = m^{4} – m^{2}

So, m^{4} – m^{2} – 72 = 0

(m^{2} – 9)(m^{2} + 8) = 0

So, m = 9 and m^{2} + 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 x^{2}/a^{2} – y^{2}/b^{2} + 1 = 0 is defined only when b > a.

Reason: Director circle of hyperbola x^{2}/25 – y^{2}/9 = 1 is x^{2} + y^{2} = 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 x^{2}/a^{2 }– y^{2}/b^{2} + 1 = 0 is

y = mx ± √(b^{2} - a^{2}m^{2})

or (y – mx)^{2} = b^{2} – a^{2}m^{2}

so, m^{2}(x^{2} + a^{2}) -2mxy + y^{2} - b^{2} = 0

since, m_{1}m_{2} = -1

So, (y^{2} - b^{2})/(x^{2} + a^{2}) = -1

or x^{2} + y^{2} = b^{2} - a^{2}, b > a

Also, director circle of x^{2}/25 – y^{2}/9 = 1 is

x^{2} + y^{2} = 25 – 9 = 16.

**Illustration 12:** C is a centre of the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 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

x^{2}/a^{2} – y^{2}/b^{2} = 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

⇒ h^{2} + k^{2} = (h–acosθ/1–sinθ)^{2} + (k–bcosθ/1–sinθ)^{2}

⇒ h^{2} + k^{2} = (h–acosθ/1–sinθ)^{2} + (k–bcosθ/1–sinθ)^{2}

⇒ h^{2} + k^{2} = h^{2} + k^{2} + (a^{2} + b^{2}) cos^{2}θ/(1–sinθ)^{2} – (2ah + 2bk) cosθ/1–sinθ

⇒ 2(ah + bk) = (a^{2} + b^{2}) cosθ/1–sinθ …… (5)

Similarly OC = OR

Hence h^{2} + k^{2} = (h–acosθ/1+sinθ)^{2} + (k+bcosθ/1+sinθ)^{2}

Which on simplification as in the last case, given 2(ah –bk) = (a^{2} + b^{2}) 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(a^{2}h^{2} – b^{2}k^{2}) = (a^{2} + b^{2}) cos^{2}θ/1–sin^{2}θ = (a^{2} + b^{2})

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

4(a^{2}x^{2} – b^{2}y^{2}) = (a^{2} + b^{2})^{2}

**Illustration 13:** The equation of a hyperbola conjugate to the hyperbola x^{2} + 3xy + 2y^{2} + 2x + 3y = 0 is

(a) x^{2} + 3xy + 2y^{2} + 2x + 3y + 1 = 0

(b) x^{2} + 3xy + 2y^{2} + 2x + 3y + 2 = 0

(c) x^{2} + 3xy + 2y^{2} + 2x + 3y + 3 = 0

(d) x^{2} + 3xy + 2y^{2} + 2x + 3y + 4 = 0

**Solution: **Since, the given hyperbola is H : x^{2} + 3xy + 2y^{2} + 2x + 3y = 0

Let the pair of asymptotes be x^{2} + 3xy + 2y^{2} + 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 : x^{2} + 3xy + 2y^{2} + 2x + 3y + 1 = 0

since, H + C = 2A

So, C = 2A – H

= x^{2} + 3xy + 2y^{2} + 2x + 3y + 2

Hence, the conjugate hyperbola is x^{2} + 3xy + 2y^{2} + 2x + 3y + 2 = 0

**Illustration 14: **The locus of the point of intersection of two perpendicular tangents to the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 is

(a) director circle

(b) x^{2} + y^{2} = a^{2}

(c) x^{2} + y^{2} = a^{2} - b^{2}

(d) x^{2} + y^{2} = a^{2} + b^{2}

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

y = mx + (a^{2}m^{2} – b^{2})

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

(k - mh)^{2} = a^{2}m^{2} – b^{2}

So, m^{2}(h^{2} - a^{2}) – 2mhk + k^{2} + b^{2} = 0

This is a quadratic in m

Let the slopes of tangents be m_{1} and m_{2}.

then m_{1}m_{2 }= -1.

So, (k^{2} + b^{2})/(h^{2} – a^{2}) = -1

(h^{2} + k^{2}) = (a^{2} – b^{2})

Hence, the locus is (x^{2} + y^{2}) = (a^{2} – b^{2}) which is the director circle of x^{2}/a^{2} – y^{2}/b^{2} = 1.

**Illustration 15: **If a tangent to the hyperbola x^{2}/25 – y^{2}/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, λ^{2} = 25 cos^{2}θ = 16 cot^{2}θ

So, sec^{2}θ = 25/ λ^{2} and tan^{2}θ = 16/λ^{2}

So, sec^{2}θ – tan^{2}θ = 25/λ^{2} – 16/λ^{2} =1

so, λ^{2} = 25 - 16 = 9

So, λ = 3.

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