the conic below and find its foci. 36x2 - 25y2 - 288x + 150y + 1251 = 0 SOLUTION We complete the squares as follows. 25(y? - 6y) - 36(x2 - 8x) = 1251 25(7 - 6y + [ 8x + = 1251 + 225 - 576 25(y-( ) - 36(x - (v-C = 1 This is in the standard form except that x and y are replaced by x - 4 and y - 3. Thus a? = I, b2 = ], and c2 = . The hyperbola is shifted four units to the right and three units upward. The foci are (x, y) = | and (4, 3 - V61) and the vertices are (x, y) = and (4, -3). The asymptotes are y - 3 = +(x - 4). The hyperbola is sketched in the figure.
the conic below and find its foci. 36x2 - 25y2 - 288x + 150y + 1251 = 0 SOLUTION We complete the squares as follows. 25(y? - 6y) - 36(x2 - 8x) = 1251 25(7 - 6y + [ 8x + = 1251 + 225 - 576 25(y-( ) - 36(x - (v-C = 1 This is in the standard form except that x and y are replaced by x - 4 and y - 3. Thus a? = I, b2 = ], and c2 = . The hyperbola is shifted four units to the right and three units upward. The foci are (x, y) = | and (4, 3 - V61) and the vertices are (x, y) = and (4, -3). The asymptotes are y - 3 = +(x - 4). The hyperbola is sketched in the figure.
Chapter11: Conics
Section11.2: Parabolas
Problem 11.38TI: Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the...
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Transcribed Image Text:EXAMPLE 7
Sketch the conic below and find its foci.
36x2 - 25y2 - 288x + 150y + 1251 = 0
SOLUTION We complete the squares as follows.
25(y? - 6y) – 36(x2 - 8x) = 1251
25(v²– 6
- бу +
- 8x +
= 1251 + 225 - 576
25( y -
=|
3D1
This is in the standard form except that x and y are replaced by x - 4 and y - 3. Thus a? =
b2 =
I, and c? =
The hyperbola is shifted four units to
the right and three units upward. The foci are (x, y) =
and (4, 3 - V61) and the vertices are (x, y) =
and (4, -3). The asymptotes are
y - 3 = +
- 4). The hyperbola is sketched in the figure.
Transcribed Image Text:EXAMPLE 3
Find the equation of the ellipse with foci (0, ±6) and vertices (0, +7).
SOLUTION Using the standard notation, we have c =
and a =
. Then we obtain b2 = a? - 2 = 49
|, so an equation of the ellipse is
1.
Another way of writing this equation is 49x2 + 13y2 |
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