+0

# find the focus of x^2-16x+y+68=0

0
467
1

find the focus and directrix of x^2-16x+y+68=0

Jan 5, 2015

#1
+5

x² - 16x + y + 68  =  0

--->  y  =  -x² + 16x - 68

Complete the square:

--->  y + 68  =  -x² + 16 x

Factor out the negative sign:

--->  y + 68  =  -(x² - 16 x)

Complete the square by dividing the linear term by 2, squaring the rsult and adding that result to both sides:

--->  y + 68 - 64  =  -(x² - 16x + 64)

--->  y + 4  =  -(x - 8)²

The vertex is:  (8, -4)

The focal length is found using:  f  =  1/(4a)   --->   f  =  1/(4·-1)  =  -1/4

The focal point will be at (8, -4 - 1/4)  =  (8, -4 1/4)

The directrix will be:  y  =  -4 + 1/4   --->   y  =  -3 1/4

Jan 5, 2015

#1
+5

x² - 16x + y + 68  =  0

--->  y  =  -x² + 16x - 68

Complete the square:

--->  y + 68  =  -x² + 16 x

Factor out the negative sign:

--->  y + 68  =  -(x² - 16 x)

Complete the square by dividing the linear term by 2, squaring the rsult and adding that result to both sides:

--->  y + 68 - 64  =  -(x² - 16x + 64)

--->  y + 4  =  -(x - 8)²

The vertex is:  (8, -4)

The focal length is found using:  f  =  1/(4a)   --->   f  =  1/(4·-1)  =  -1/4

The focal point will be at (8, -4 - 1/4)  =  (8, -4 1/4)

The directrix will be:  y  =  -4 + 1/4   --->   y  =  -3 1/4

geno3141 Jan 5, 2015