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′ = 2a . 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 = ae 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 SP2 = e2PM2.
Then (x - ae)2 + ( y - 0)2 = e2 (
x – a/e )2
Þ (x - ae)2 + y2 = (ex - a)2
Þ (e2 -1)x2
- y2 = a2 (e2
-1)
= 1. Since e > 1, a2 (e2
-1) > 0 . Setting a2(e2
-1) = b2, we obtain the locus of P as x2/a2 - y2/b2
= 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 b2 = c2
- a2 .
(1) The line segment AA′ is the transverse axis of length 2a .
(2) The line segment BB′ is the conjugate axis of length 2b .
(3) The line segment CA = the line
segment CA′ = semi transverse axis = a and the line segment CB
= the line segment CB′ = semi conjugate axis = b .
(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 2b2 / 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 x2 + y2
= a2.
(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¢ | = 2a . (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)
⇔ y2
= - 4√2x.
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)
⇔ y2 + 4 y + 4
= -12x + 60
⇔ y2 + 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 = 4a
( y + 2) .
Since this passes through (3,6), we get
(3 +1)2 = 4a
(6 + 2)
⇔ a = 1/2.
Then the equation of parabola is ( x +1)2 = 2 ( y + 2)
which on simplifying yields,
x2 + 2x - 2y - 3 = 0 .
Example 5.17
Find the vertex, focus, directrix, and length of the latus rectum
of the parabola x2 - 4x - 5 y -1 = 0.
Solution
For the parabola,
x2 - 4x - 5y -1 = 0
⇔ x2 - 4x = 5y +1
⇔ x2 - 4x + 4 = 5y +1+ 4 .
⇒ ( x - 2)2 =
5( y +1) which is in standard form. Therefore, 4a = 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
4y +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′ = 2c and 2c = 4; A′A =
2a = 6
⇒ c =
2 and a = 3 ,
⇒ b2
= a2 − c2 = 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 x2/9 + y2/5
= 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 SP2
= e2 PM 2.
Therefore, the length of major axis = 2a = and
the length of minor axis = 2b =
Example 5.20
Find the foci, vertices and length of major and minor axis of
the conic
4x2 + 36 y2 + 40x −
288 y + 532 = 0 .
Solution
Completing the square on x
and y of 4x2 + 36y2 + 40x - 288y + 532 = 0 ,
4(x2 +10x + 25 - 25) + 36( y2
- 8 y +16 -16) + 532 = 0 , gives
4(x2 +10x + 25 ) + 36( y2
- 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, c2 = a2 − b2
= 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 = 2a
= 12 units and
the length of the minor axis = 2b = 4 units.
Example 5.21
For the ellipse 4x2 + y2 + 24x
− 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
4x2 + 24x + y2 − 2 y
+ 21 = 0
That is, 4 (x2 + 6x + 9 −
9) + ( y2 − 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
c2 = 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 = 2b2 /
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¢ = 2c = 12, c = 6
a = 4
b2 = c2
- a2 = 36 -16 = 20 .
Hence the equation of the required hyperbola is y2/16 - x2/20 =
1 =
Example 5.23
Find the vertices, foci for the hyperbola 9x2
−16y2 = 144 .
Solution
Reducing 9x2 −16 y2= 144 to
the standard form,
We have,
With the transverse axis is along x -axis vertices are
(−4, 0) and (4, 0) ;
and c2 = a2 + b2
= 16 + 9 = 25 , c = 5 .
Hence the foci are (−5, 0) and (5, 0) .
Find the centre, foci, and eccentricity of the hyperbola 11x2
− 25 y2 − 44x + 50 y − 256 = 0
Rearranging terms in the equation of hyperbola to bring it to
standard form,
we have, 11(x2 − 4x) − 25( y2
− 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) , a2 = 25, b2
= 11
c2 = a2 + b2
= 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 2a = 36.18, 2b = 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.
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