Examples On Propositions Of A Hyperbola

Hyperbola is defined as a conic with an eccentricity greater than 1 [e > 1]. It is the locus of a point where the distance from a fixed point called focus and from the line which is fixed called directrix is a constant called eccentricity [denoted as “e”].

x2/ a2 – y2/ b2 = 1 where b2 = a2 (e2 – 1)

is the standard equation of a hyperbola. The equation of the hyperbola with focus (h, k), directrix lx + my + n = 0 and eccentricity ‘e’ is

(x – h)2 + (y – k)2 = e2 (lx + my + n)2/(l2 + m2)

Propositions Of A Hyperbola

The properties and propositions of the hyperbola are as follows:

  • The diameter of a hyperbola is the locus of the centre point of a group of parallel chords. It is given by y = b2 x / (a2 m), m being the slope of the group of parallel chords.
  • The bisection of the chords which are parallel to each other by the two diameters of the hyperbola are called conjugate diameters. It is given by y = mx and y = m1x of the hyperbola x2/ a2 – y2/ b2 = 1 are conjugate if mm1 = b2 / a2.
  • The latus rectum is a chord through one of the foci and at a right angle to the transverse axis of a hyperbola. The length of the latus rectum is given by 2b2 /a2.
  • The locus of the point of intersection of a pair of perpendicular tangents to a hyperbola is the director circle. It is given by x2 / a2 – y2 / b2 = 1 is x2 + y2 = a2 – b2
  • A line which is a tangent to the hyperbola at a point of infinity but not infinity by itself can be the asymptote of the curve.

Propositions Of A Hyperbola Solved Examples for IIT JEE

Example 1: What is the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13?

Solution:

Conjugate axis is 5 and the distance between foci = 13

2b = 5 and 2ae = 13.

b2 = a2 (e2 − 1)

25 / 4 = [(13)2 / 4e2] * (e2 − 1)

25 / 4 = [169 / 4] − [169 / 4e2] or e2 = 169 / 144

e = 13 / 12

or a = 6, b = 5 / 2 or hyperbola is x2 / 36 − y2 / [25 / 4] = 1

25 x2 − 144 y2 = 900

Example 2: A hyperbola passes through the points (3, 2) and (17, 12) and has its centre at the origin and transverse axis is along the x-axis. Find the length of its transverse axis.

Solution:

Let the equation of hyperbola is x2 / a2 – y2 / b2 = 1

But it passes through (3, 2)

9 / a2 − 4 / b2 = 1 ….. (i)

Also its passes through (17, 12)

(−17)2 / a2 − (12)2 / b2 = 1 ….. (ii)

Solving these, we get a = 1 and b = √2

Hence, length of transverse axis = 2a = 2

Example 3: The distance between the foci of a hyperbola is double the distance between its vertices and the length of its conjugate axis is 6. Find the equation of the hyperbola referred to its axes as axes of co-ordinates.

Solution:

According to given conditions, 2ae = 2 * 2a or

√[(x − 2)2 + y2] = 4 − √[(x − 2)2 + y2] and

A = (0, 0) ; B = (4a, 4a)

Hence, a = 3 / √3 = √3

Therefore, equation is x2 / 3 − y2 / 9 = 1

i.e., 3x2 − y2 = 9

Example 4: What are the equation of the directrices of the conic x2 + 2x − y2 + 5 = 0?

Solution:

(x + 1)2 − y2 −1 + 5 = 0

− [(x + 1)2 / 4] + y2 / 4 = 1

Equation of directrices of y2 / b2 − x2 / a2 = 1 are y = ± be

Here b = 2, e = √1 + 1 = √2

Hence, y = ± 2 / √2

y = ± √2

Example 5: The value of m for which y = mx + 6 is a tangent to the hyperbola x2 / 100 − y2 / 49 = 1, is ______________.

Solution:

If y = mx + c touches x2 / a2 – y2 / b2 = 1, then c2 = a2m2 −b2.

Here c = 6, a2 = 100, b2 = 49

∴ 36 = 100m2 − 49

⇒ 100m2 = 85

⇒ m = √[17 / 20]

Example 6: Find the equation of axis of the given hyperbola x23y22=1\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1 which is equally inclined to the axes 

A) y = x + 1                                   

B) y = x − 1

C) y = x + 2                                  

D) y = x − 2

Solution:

x23y22=1\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1

Equation of tangent are equally inclined to the axis i.e.,  

tanθ=1=m\tan \theta =1=m

Equation of tangent y=mx+a2m2b2y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}

Given equation is x23y22=1\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1 is a equation of hyperbola which is of form x2a2y2b2=1\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1

Now, on comparing

a2=3,b2=2y=1.x+3×(1)22y=x+1{{a}^{2}}=3, {{b}^{2}}=2 \\ y=1.x+\sqrt{3\times {{(1)}^{2}}-2}\\ y=x+1

Example 7: If q is the acute angle of intersection at a real point of intersection of the circle x2+y2=5{{x}^{2}}+{{y}^{2}}=5 and the parabola y2=4x{{y}^{2}}=4x then find the value of tanq.

Solution: 

Solving equations

x2+y2=5 and y2=4x{{x}^{2}}+{{y}^{2}}=5 \text \ and \ {{y}^{2}}=4x we get

x2+4x5=0x=1,5{{x}^{2}}+4x-5=0 \Rightarrow x=1,\,-5

For x = 1, y2=4y=±2{{y}^{2}}=4\\ y=\pm 2 

For x = -5, y2=20{{y}^{2}}=-20 (imaginary values) 

The points are (1, 2)(1, 2).

m1{{m}_{1}} for x2+y2=5{{x}^{2}}+{{y}^{2}}=5 at (1, 2) 

dydx=xy(1,2)=12{{\left. \frac{dy}{dx}=-\frac{x}{y} \right|}_{(1,\,2)}}=-\frac{1}{2}

Similarly m2{{m}_{2}} for y2=4x{{y}^{2}}=4x at (1, 2) is 1.

tanθ=m1m21+m1m2=121112=3\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{-\frac{1}{2}-1}{1-\frac{1}{2}} \right|=3

Example 8: If the foci of the ellipse x216+y2b2=1\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 and the hyperbola x2144y281=125\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25} coincide, then find the value of b2{{b}^{2}}.

Solution:

Hyperbola is 

x2144y281=125a=14425,b=8125,e1=1+81144=225144=1512=54\frac{{{x}^{2}}}{144}-\frac{{{y}^{2}}}{81}=\frac{1}{25}\\ a=\sqrt{\frac{144}{25}},\,\,b=\sqrt{\frac{81}{25}},\,\,{{e}_{1}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}

Therefore, foci =(ae1,0)=(125.54,0)=(3,0)=(a{{e}_{1}},0)=\left( \frac{12}{5}.\frac{5}{4},0 \right)=(3,\,0)

Therefore, focus of ellipse =(4e,0)i.e. (3,0)=(4e,0) \text i.e. \ (3,\,0)

Hence, b2=16(1916)=7{{b}^{2}}=16\left( 1-\frac{9}{16} \right)=7  

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