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1.5 Conic sections  (Page 6/23)

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− 2 c x = 4 a 2 + 4 a ( x + c ) 2 + y 2 + 2 c x 4 a ( x + c ) 2 + y 2 = −4 a 2 − 4 c x ( x + c ) 2 + y 2 = − a − c x a ( x + c ) 2 + y 2 = a 2 + 2 c x + c 2 x 2 a 2 x 2 + 2 c x + c 2 + y 2 = a 2 + 2 c x + c 2 x 2 a 2 x 2 + c 2 + y 2 = a 2 + c 2 x 2 a 2 .

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

x 2 − c 2 x 2 a 2 + y 2 = a 2 − c 2 ( a 2 − c 2 ) x 2 a 2 + y 2 = a 2 − c 2 .

Finally, divide both sides by a 2 − c 2 . This gives the equation

x 2 a 2 + y 2 a 2 − c 2 = 1 .

We now define b so that b 2 = c 2 − a 2 . This is possible because c > a . Therefore the equation of the ellipse becomes

x 2 a 2 − y 2 b 2 = 1 .

Finally, if the center of the hyperbola is moved from the origin to the point ( h , k ) , we have the following standard form of a hyperbola.

Equation of a hyperbola in standard form

Consider the hyperbola with center ( h , k ) , a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1

and the foci are located at ( h ± c , k ) , where c 2 = a 2 + b 2 . The equations of the asymptotes are given by y = k ± b a ( x − h ) . The equations of the directrices are

x = k ± a 2 a 2 + b 2 = h ± a 2 c .

If the major axis is vertical, then the equation of the hyperbola becomes

( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1

and the foci are located at ( h , k ± c ) , where c 2 = a 2 + b 2 . The equations of the asymptotes are given by y = k ± a b ( x − h ) . The equations of the directrices are

y = k ± a 2 a 2 + b 2 = k ± a 2 c .

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form A x 2 + B y 2 + C x + D y + E = 0 , where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.

Finding the standard form of a hyperbola

Put the equation 9 x 2 − 16 y 2 + 36 x + 32 y − 124 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

First add 124 to both sides of the equation:

9 x 2 − 16 y 2 + 36 x + 32 y = 124 .

Next group the x terms together and the y terms together, then factor out the common factors:

( 9 x 2 + 36 x ) − ( 16 y 2 − 32 y ) = 124 9 ( x 2 + 4 x ) − 16 ( y 2 − 2 y ) = 124.

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In the first set of parentheses, take half the coefficient of x and square it. This gives ( 4 2 ) 2 = 4 . In the second set of parentheses, take half the coefficient of y and square it. This gives ( −2 2 ) 2 = 1 . Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

9 ( x 2 + 4 x + 4 ) − 16 ( y 2 − 2 y + 1 ) = 124 + 36 − 16 9 ( x 2 + 4 x + 4 ) − 16 ( y 2 − 2 y + 1 ) = 144.

Next factor both sets of parentheses and divide by 144:

9 ( x + 2 ) 2 − 16 ( y − 1 ) 2 = 144 9 ( x + 2 ) 2 144 − 16 ( y − 1 ) 2 144 = 1 ( x + 2 ) 2 16 − ( y − 1 ) 2 9 = 1.

The equation is now in standard form. Comparing this to [link] gives h = −2 , k = 1 , a = 4 , and b = 3 . This is a horizontal hyperbola with center at ( −2 , 1 ) and asymptotes given by the equations y = 1 ± 3 4 ( x + 2 ) . The graph of this hyperbola appears in the following figure.

A hyperbola is drawn with equation 9x2 + 16y2 + 36x + 32y – 124 = 0. It has center at (−2, 1), and the hyperbolas are open to the left and right.
Graph of the hyperbola in [link] .
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Put the equation 4 y 2 − 9 x 2 + 16 y + 18 x − 29 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

( y + 2 ) 2 9 − ( x − 1 ) 2 4 = 1 . This is a vertical hyperbola. Asymptotes y = −2 ± 3 2 ( x − 1 ) .
A hyperbola is drawn with equation 4y2 – 9x2 + 16x + 18y – 29 = 0. It has center at (1, −2), and the hyperbolas are open to the top and bottom.

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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