Question -
Answer -
(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity= 2
Given:
Focus = (0, 3)
Directrix => x + y – 1 = 0
Eccentricity = 2
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
So, 2{x2 + y2 + 9 – 6y} =4{x2 + y2 + 1 – 2x – 2y + 2xy}
2x2 + 2y2 + 18 – 12y = 4x2 +4y2+ 4 – 8x – 8y + 8xy
2x2 + 2y2 + 18 – 12y – 4x2 –4y2 – 4 – 8x + 8y – 8xy = 0
– 2x2 – 2y2 – 8x – 4y – 8xy+ 14 = 0
-2(x2 + y2 – 4x + 2y + 4xy –7) = 0
x2 + y2 – 4x + 2y + 4xy – 7= 0
∴The equation of hyperbola is x2 + y2 –4x + 2y + 4xy – 7 = 0
(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 andeccentricity = 2
Focus = (1, 1)
Directrix => 3x + 4y + 8 = 0
Eccentricity = 2
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
25{x2 + 1 – 2x + y2 + 1 –2y} = 4{9x2 + 16y2+ 64 + 24xy + 64y + 48x}
25x2 + 25 – 50x + 25y2 + 25– 50y = 36x2 + 64y2 + 256 + 96xy + 256y + 192x
25x2 + 25 – 50x + 25y2 + 25– 50y – 36x2 – 64y2 – 256 – 96xy – 256y –192x = 0
– 11x2 – 39y2 – 242x – 306y– 96xy – 206 = 0
11x2 + 96xy + 39y2 + 242x +306y + 206 = 0
∴The equation of hyperbola is11x2 + 96xy +39y2 + 242x + 306y + 206 = 0
(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3
Given:
Focus = (1, 1)
Directrix => 2x + y = 1
Eccentricity =√3
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
5{x2 + 1 – 2x + y2 + 1 – 2y}= 3{4x2 + y2+ 1 + 4xy – 2y – 4x}
5x2 + 5 – 10x + 5y2 + 5 –10y = 12x2 + 3y2 + 3 + 12xy – 6y – 12x
5x2 + 5 – 10x + 5y2 + 5 –10y – 12x2 – 3y2 – 3 – 12xy + 6y + 12x = 0
– 7x2 + 2y2 + 2x – 4y – 12xy+ 7 = 0
7x2 + 12xy – 2y2 – 2x + 4y–7 = 0
∴The equation of hyperbola is7x2 + 12xy – 2y2 –2x + 4y– 7 = 0
(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity= 2
Given:
Focus = (2, -1)
Directrix => 2x + 3y = 1
Eccentricity = 2
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
13{x2 + 4 – 4x + y2 + 1 +2y} = 4{4x2 + 9y2 + 1 + 12xy – 6y – 4x}
13x2 + 52 – 52x + 13y2 + 13+ 26y = 16x2 + 36y2 + 4 + 48xy – 24y – 16x
13x2 + 52 – 52x + 13y2 + 13+ 26y – 16x2 – 36y2 – 4 – 48xy + 24y +16x = 0
– 3x2 – 23y2 – 36x + 50y –48xy + 61 = 0
3x2 + 23y2 + 48xy + 36x –50y– 61 = 0
∴The equation of hyperbola is3x2 + 23y2 +48xy + 36x – 50y– 61 = 0
(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity =2
Given:
Focus = (a, 0)
Directrix => 2x + 3y = 1
Eccentricity = 2
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
45{x2 + a2 – 2ax + y2}= 16{4x2 + y2 + a2 – 4xy – 2ay+ 4ax}
45x2 + 45a2 – 90ax + 45y2 =64x2 + 16y2 + 16a2 – 64xy –32ay + 64ax
45x2 + 45a2 – 90ax + 45y2 –64x2 – 16y2 – 16a2 + 64xy +32ay – 64ax = 0
19x2 – 29y2 + 154ax – 32ay –64xy – 29a2 = 0
∴The equation of hyperbola is19x2 – 29y2 +154ax – 32ay – 64xy – 29a2 = 0
(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2
Given:
Focus = (2, 2)
Directrix => x + y = 9
Eccentricity = 2
Now, let us find the equation of the hyperbola
Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that (a – b)2 =a2 + b2 + 2ab &(a + b + c)2 =a2 + b2 + c2 + 2ab + 2bc + 2ac]
x2 + 4 – 4x + y2 + 4 – 4y =2{x2 + y2 + 81 + 2xy – 18y – 18x}
x2 – 4x + y2 + 8 – 4y = 2x2 +2y2 + 162 + 4xy – 36y – 36x
x2 – 4x + y2 + 8 – 4y – 2x2 –2y2 – 162 – 4xy + 36y + 36x = 0
– x2 – y2 + 32x + 32y + 4xy– 154 = 0
x2 + 4xy + y2 – 32x – 32y +154 = 0
∴The equation of hyperbola isx2 + 4xy + y2 –32x – 32y + 154 = 0