Find Vertex Focus Equation of Directrix of Hyperbola :
Here we are going to see some example problems find vertex focus equation of directrix of hyperbola.
Symmetric about x- axis Center : C (0, 0) Foci : F1 (ae, 0) F2 (-ae, 0) Vertices on transverse axis : A (a, 0) A' (-a, 0) Vertices on conjugate axis : B (0, b) B' (0, -b) Equation of directrices : x = ± (a/e) |
Symmetric about y- axis Center : C (0, 0) Foci : F1 (0, ae) F2 (0, -ae) Vertices on transverse axis : A (0, a) A' (0, -a) Vertices on conjugate axis : B (b, 0) B' (-b, 0) Equation of directrices : y = ± (a/e) |
Symmetric about x- axis Center : C (h, k) Foci : F1 (h + c, k ) F2 (h - c, k ) Vertices on major axis : A (h + a, k) A' (h - a, k) Equation of directrices : x = h ± (a/e) |
Symmetric about y- axis Center : C (h, k) Foci : F1 (h, k + c) F2 (h, k - c) Vertices on major axis : A (h, k + a ) A' (h, k - a) Equation of directrices : y = k ± (a/e) |
Note :
e = √[1 + (b2/a2)]
b2 = a2(e2 - 1)
Let us look into the next problem on "Find Vertex Focus Equation of Directrix of Hyperbola".
Question 1 :
Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:
(i) [(x + 3)2/225] - [(y - 4)2/64] = 1
Solution :
The given conic represents the " Hyperbola "
The given ellipse is symmetric about x - axis.
a2 = 225 and b2 = 64
a = 15 and b = 8
c2 = a2 + b2
c2 = 225 + 64
c2 = 289
c = 17
e = c/a = 17/15
Center :
C (h, k) ==> (-3, 4)
Foci :
F1 (h + c, k ) F2 (h - c, k )
F1 (-3 + 17, 4) F2 (-3 - 17, 4 )
F1 (14, 4) F2 (-20, 4 )
Vertices on transverse axis :
A (h + a, k) A' (h - a, k)
A (-3 + 15, 4) A' (-3 - 15, 4)
A (12, 4) A' (-18, 4)
Equation of directrices :
x = h ± (a/e)
x = -3 ± (15/(17/15))
x = -3 ± (225/17)
x = -3 + (225/17) x = (-51 + 225)/17 x = 174/17 |
x = -3 - (225/17) x = (-51 - 225)/17 x = -276/17 |
Let us look into the next problem on "Find Vertex Focus Equation of Directrix of Hyperbola".
(ii) [(y - 2)2/25] - [(x + 1)2/16] = 1
Solution :
The given conic represents the " Hyperbola "
The given ellipse is symmetric about y - axis.
a2 = 25 and b2 = 16
a = 5 and b = 4
c2 = a2 + b2
c2 = 25 + 16
c2 = 41
c = √41
e = c/a = √41/5
Center :
C (h, k) ==> (-1, 2)
Foci :
F1 (h, k + c) F2 (h, k - c)
F1 (-1, 2 + √41) F2 (-1, 2 - √41)
Vertices on transverse axis :
A (h, k + a) A' (h, k - a)
A (-1, 2 + 5) A' (-1, 2 - 5)
A (-1, 7) A' (-1, -3)
Equation of directrices :
y = k ± (a/e)
y = 2 ± (5/(√41/5))
y = 2 ± (25/√41)
y = 2 + (25/√41) |
y = 2 - (25/√41) |
Let us look into the next problem on "Find Vertex Focus Equation of Directrix of Hyperbola".
(v) 18x2 + 12y2 − 144x + 48y + 120 = 0
Solution :
18x2 + 12y2 − 144x + 48y + 120 = 0
18x2 − 144x + 12y2 + 48y + 120 = 0
18(x2 - 8x) + 12(y2 + 4y) + 120 = 0
18[(x - 4)2 - 16] + 12[(y + 2)2 - 4] + 120 = 0
18(x - 4)2 - 288 + 12(y + 2)2 - 48 + 120 = 0
18(x - 4)2 + 12(y + 2)2 - 336 + 120 = 0
18(x - 4)2 + 12(y + 2)2 - 216 = 0
18(x - 4)2 + 12(y + 2)2 = 216
[(x - 4)2/12] + [(y + 2)2/18] = 1
The given conic represents the " Ellipse "
The given ellipse is symmetric about y - axis.
a2 = 18 and b2 = 12
a = 3√2 and b = 2√3
c2 = a2 - b2
c2 = 18 - 12
c2 = 6
c = √6
e = c/a = √6/3√2
e = 1/√3
Center :
C (h, k) ==> (4, -2)
Foci :
F1 (h, k - c) F2 (h, k + c)
F1 (4, -2 - √6) , F2 (4, -2 + √6)
Vertices on major axis :
A (h, k - a) A' (h, k + a)
A (4, -2-3√2) A' (4, -2 + 3√2)
Equation of directrices :
y = k ± (a/e)
y = -2 ± (3√2/(1/√3))
y = -2 ± 3√6
y = -2 + 3√6 |
y = -2 - 3√6 |
Let us look into the next problem on "Find Vertex Focus Equation of Directrix of Hyperbola".
(vi) 9x2 − y2 − 36x − 6y + 18 = 0
Solution :
9x2 − y2 − 36x − 6y + 18 = 0
9x2− 36x − y2 − 6y + 18 = 0
9[x2− 4x] − [y2 + 6y] + 18 = 0
9[(x - 2)2− 4] − [(y + 3)2 - 9] + 18 = 0
9(x - 2)2− 36 − (y + 3)2 + 9 + 18 = 0
9(x - 2)2 - (y + 3)2 = 9
[(x - 2)2/1] - [(y + 3)2/9] = 1
The given conic represents the " Hyperbola "
The given ellipse is symmetric about x - axis.
a2 = 1 and b2 = 9
a = 1 and b = 3
c2 = a2 + b2
c2 = 1 + 9
c2 = 10
c = √10
e = c/a = √10/1
e = √10
Center :
C (h, k) ==> (2, -3)
Foci :
F1 (h + c, k ) F2 (h - c, k )
F1 (2 + √10, -3) F2 (2 - √10, -3)
Vertices on transverse axis :
A (h + a, k) A' (h - a, k)
A (2 + 1, -3) A' (2 - 1, -3 )
A (3, -3) A' (1, -3)
Equation of directrices :
x = h ± (a/e)
x = 2 ± (1/√10)
x = 2 + (1/√10) |
x = 2 - (1/√10) |
After having gone through the stuff given above, we hope that the students would have understood, "Find Vertex Focus Equation of Directrix of Hyperbola".
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