We invoke that a hyperbola is the locus of a point which moves such that its distance from a fixed point (focus) bears a constant ratio (eccentricity) greater than unity its distance from its directrix, bearing a constant ratio e (e > 1) .

**Hyperbola**

We invoke that a hyperbola is the locus of a point which moves
such that its distance from a fixed point (focus) bears a constant ratio
(eccentricity) greater than unity its distance from its directrix, bearing a
constant ratio *e *(*e *> 1) .

Let *S *be a focus, *l *be the directrix line, *e *be
the eccentricity *e *> 1 and *P*(*x*, *y*) be the moving
point. Draw *SZ *and *PM *perpendicular to *l .*

Let *A *and *A*′ be the points which divide *SZ *internally
and externally in the ratio *e *:1 respectively.

Let *AA*′ = 2*a *. Let the point of intersection of the
perpendicular bisector with *AA*′ be *C *. Then CA= CA*' *= *a
*Choose *C *as origin and the line *CZ *produced as *x *-axis
and the perpendicular bisector of *AA*′ as *y *-axis.

Þ AS = eAZ A¢S = eA¢Z

Þ CS - CA = e(CA - CZ ) A¢C + CS = e( A¢C + CZ )

Þ CS - a = e(a - CZ ) …
(1) a + CS = e(a + CZ ) …
(2)

(1) + (2)
gives CS = *a*e and (2) - (1) gives CZ = a/e.

Hence, the coordinates of S are (ae, 0) . Since PM = *x* – *a*/e
, the equation of directrix is *x* – *a*/e = 0 . Let P(x, y) be any point on
the hyperbola.

By the definition of a conic, SP/ PM = e

Þ SP^{2} = e^{2}PM^{2}.

Then (x - ae)^{2} + ( y - 0)^{2} = e^{2} (
x – *a*/e )^{2}

Þ (x - *a*e)^{2} + y^{2} = (ex - *a*)^{2}

Þ (e^{2} -1)x^{2}
- y^{2} = *a*^{2} (e^{2}
-1)

= 1. Since e > 1, a^{2 }(e^{2}
-1) > 0 . Setting a^{2}(e^{2}
-1) = b^{2}, we obtain the locus of P as x^{2}/a^{2} - y^{2}/b^{2}
= 1 = which is the equation of a **Hyperbola
in standard form** and note that it is symmetrical about *x *and *y*-axes.

Taking *ae *= *c *, we get *b*^{2} = *c*^{2}
- *a*^{2} .

(1) The line segment *AA*′ is the **transverse **axis of length 2*a *.

(2) The line segment *BB*′ is the **conjugate **axis of length 2*b *.

(3) The line segment *CA *= the line
segment *CA*′ = **semi transverse axis **= ** a **and the line segment

(4) By symmetry, taking *S*′(−*ae*,
0) as focus and *x *=− *a/e *as directrix *l*′ gives the same
hyperbola.

Thus we see that a hyperbola has two foci *S *(*ae*,
0) and *S*¢(-*ae*, 0) , two
vertices *A*(*a*, 0)

and *A*¢(-*a*, 0) and two
directrices *x *= *a/e *and *x *=- *a/e*.

Length of latus rectum of hyperbola is 2b^{2} / *a*, which can be obtained along lines as
that of the ellipse.

**Asymptotes**

Let *P*(*x*, *y*) be a point on the curve defined
by *y *= *f *(*x*) , which moves further and further away from
the origin such that the distance between *P *and some fixed line tends to
zero. This fixed line is called an **asymptote**.

Note that the hyperbolas admit asymptotes while parabolas and
ellipses do not.

(1) The circle described on the transverse axis of hyperbola as
its diameter is called the auxiliary circle of the hyperbola. Its equation is *x*^{2} + *y*^{2}
= *a*^{2}.

(2) The absolute difference of the focal distances of any point
on the hyperbola is constant and is equal to length of transverse axis. That is, | *PS *- *PS*¢ | = 2*a *. (can be proved similar that
of ellipse)

So far we have discussed four standard types of parabolas, two
types of ellipses and two types of hyperbolas. There are plenty of parabolas, ellipses and
hyperbolas whose equations cannot be classified under the standard types, For
instance consider the following parabola, ellipse, and hyperbola.

By a suitable transformation of coordinate axes they can be
represented by standard equations.

**Example 5.14**

Find the equation of the parabola with focus (− √2, 0) and
directrix x = √2.

**Solution**

Parabola is open left and axis of symmetry as *x*-axis and vertex (0, 0).

Then the equation of the required parabola is

(*y* - 0)^{2} = -
4√2 ( *x* - 0)

⇔ *y*^{2}
= - 4√2*x*.

**Example 5.15**

Find the equation of the parabola whose vertex is (5, -2) and
focus (2, -2).

**Solution**

Given vertex *A*(5, -2)
and focus *S*(2, -2) and the focal
distance AS = *a* = 3 .

Parabola is open left and symmetric about the line parallel to x
-axis.

Then, the equation of the required parabola is

( y + 2)^{2} = -4 (3)( x - 5)

⇔ y^{2} + 4 y + 4
= -12x + 60

⇔ y^{2} + 4 y
+12x - 56 = 0 .

**Example 5.16**

Find the equation of the parabola with vertex (-1, -2) , axis
parallel to y -axis and passing through (3, 6) .

**Solution**

Since axis is parallel to *y*
-axis the required equation of the parabola is

( x +1)^{2} = 4*a*
( *y* + 2) .

Since this passes through (3,6), we get

(3 +1)^{2} = 4*a*
(6 + 2)

⇔ *a* = 1/2.

Then the equation of parabola is ( x +1)^{2} = 2 ( y + 2)
which on simplifying yields,

x^{2} + 2x - 2y - 3 = 0 .

**Example 5.17**

Find the vertex, focus, directrix, and length of the latus rectum
of the parabola *x*^{2} - 4*x* - 5 *y* -1 = 0.

**Solution**

For the parabola,

*x*^{2} - 4*x* - 5*y* -1 = 0

⇔ x^{2} - 4x = 5*y* +1

⇔ x^{2} - 4*x* + 4 = 5*y* +1+ 4 .

⇒ ( x - 2)^{2} =
5( y +1) which is in standard form. Therefore, 4*a* = 5 and the vertex is (2, -1) , and the focus is ( 2, 1/4 ) .

Equation of directrix is *y +
k + a = 0*

y -1+ 5/4 = 0

4*y* +1 = 0 .

Length of latus rectum 5 units.

**Example 5.18**

Find the equation of the ellipse with foci (± 2, 0) , vertices
(± 3, 0) .

*Solution*

From Fig. 5.36, we get

*SS*′ = 2*c *and 2*c *= 4; *A*′*A *=
2*a *= 6

⇒ *c *=
2 and *a *= 3 ,

⇒ *b*^{2}
= *a*^{2} − *c*^{2} = 9 − 4 = 5 .

Major axis is along *x *-axis, since *a *> *b *.

Centre is (0, 0) and Foci are (±2, 0) .

Therefore, equation of the ellipse is *x*^{2}/9 + *y*^{2}/5
= 1

**Example 5.19**

Find the equation of the ellipse whose eccentricity is ½, one of the foci is (2,
3) and a directrix is *x *= 7 . Also find the length of the major and
minor axes of the ellipse.

**Solution**

By the definition of a conic, SP/PM= e or *SP*^{2}
= *e*^{2} *PM *^{2}.

Therefore, the length of major axis = 2*a *= and

the length of minor axis = 2*b *=

**Example 5.20**

Find the foci, vertices and length of major and minor axis of
the conic

4*x*^{2} + 36 *y*^{2} + 40*x *−
288 *y *+ 532 = 0 .

**Solution**

Completing the square on *x*
and *y* of 4x^{2} + 36y^{2} + 40x - 288y + 532 = 0 ,

4(*x*^{2} +10*x *+ 25 - 25) + 36( *y*^{2}
- 8 *y *+16 -16) + 532 = 0 , gives

4(*x*^{2} +10*x *+ 25 ) + 36( *y*^{2}
- 8 *y *+16 ) = -532 +100 + 576

4 ( *x *+ 5)^{2} + 36 ( *y *- 4)^{2}
= 144 .

Dividing both sides by 144 , the equation reduces to

This is an ellipse with centre (−5, 4) , major axis is parallel
to *x *-axis, length of major axis is 12 and length of minor axis is 4. Vertices are (1, 4) and (−11, 4)
.

Now, *c*^{2} = *a*^{2} − *b*^{2}
= 36 − 4 = 32

and c = ±4√2.

Then the foci are (-5 - 4√2, 4) and (-5 + 4√2, 4).

Length of the major axis = 2*a*
= 12 units and

the length of the minor axis = 2*b *= 4 units.

**Example 5.21**

For the ellipse 4*x*^{2} + *y*^{2} + 24*x
*− 2 *y *+ 21 = 0 , find the centre, vertices, and the foci. Also prove
that the length of latus
rectum is 2 .

**Solution**** **

Rearranging the terms, the equation of ellipse is

4*x*^{2} + 24*x *+ *y*^{2} − 2 *y
*+ 21 = 0

That is, 4 (*x*^{2} + 6*x *+ 9 −
9) + ( *y*^{2} − 2 *y *+1−1) + 21 = 0 ,

4 ( *x *+ 3)^{2} − 36 + ( *y *−1)^{2}
−1+ 21 = 0 ,

4 ( *x *+ 3)^{2} + ( *y *−1)^{2}
= 16 ,

Centre is (−3,1) *a *= 4, *b *= 2 , and the major axis
is parallel to *y *-axis

c^{2} = 16 - 4 = 12

*c* = ±2√3.

Therefore, the foci are (-3, 2√3 +1)
and (-3, -2√3+1)

Vertices are (1, ±4 +1) . That is the vertices are (1, 5)
and (1, -3) , and
the length of Latus
rectum = 2b^{2 }/
*a* = 2 units. (see Fig. 5.37)

Find the equation of the hyperbola with vertices (0, ±4) and
foci (0, ±6) .

From Fig. 5.38, the midpoint of line joining foci is the centre
C(0,0).

Transverse axis is *y *-axis

*AA**¢** = 2a **⇒** 2a* = 8,* *

SS¢ = 2*c* = 12*, c* = 6

*a* = 4

*b ^{2} = c^{2}
- a^{2} *= 36 -16 = 20 .

Hence the equation of the required hyperbola is *y*^{2}/16 - *x*^{2}/20 =
1 =

**Example 5.23**

Find the vertices, foci for the hyperbola 9*x*^{2}
−16*y*^{2} = 144 .

**Solution**

Reducing 9*x*^{2} −16 *y*^{2}= 144 to
the standard form,

We have,

With the transverse axis is along *x *-axis vertices are
(−4, 0) and (4, 0) ;

and *c*^{2} = *a*^{2} + *b*^{2}
= 16 + 9 = 25 , *c *= 5 .

Hence the foci are (−5, 0) and (5, 0) .

Find the centre, foci, and eccentricity of the hyperbola 11*x*^{2}
− 25 *y*^{2} − 44*x *+ 50 *y *− 256 = 0

Rearranging terms in the equation of hyperbola to bring it to
standard form,

we have, 11(*x*^{2} − 4*x*) − 25( *y*^{2}
− 2 *y*) − 256 = 0

11( *x *− 2)^{2} − 25( *y *−1)^{2}
= 256 − 44 + 25

11( *x *− 2)^{2} − 25( *y *−1)^{2}
= 275

Centre (2,1) , *a*^{2} = 25, *b*^{2}
= 11

*c*^{2} = *a*^{2} + *b*^{2}

= 25 +11 = 36

Therefore, *c *= ±6

and *e *= *c/a *= 6 and the coordinates of foci are
(8,1) and (-4,1) from Fig. 5.39.

The orbit of Halley’s Comet (Fig. 5.51) is an ellipse 36.18
astronomical units long and by 9.12 astronomical units wide. Find its
eccentricity.

Given that 2*a *= 36.18, 2*b *= 9.12 , we get

One astronomical unit (mean distance of Sun and earth) is 1, 49,
597,870 *km *, the semi major axis of the Earth’s orbit.

Tags : Equation, Definition, Theorem, Proof, Types, Solved Example Problems, Solution , 12th Mathematics : Two Dimensional Analytical Geometry II

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

12th Mathematics : Two Dimensional Analytical Geometry II : Hyperbola | Equation, Definition, Theorem, Proof, Types, Solved Example Problems, Solution

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.