Question #30429

2 Answers
May 9, 2017

Answer:

The standard form is #(x+1)^2/(sqrt5)^2+(y-1)^2/4^2=1#

Explanation:

Let's rearrange the equation by completing the squares

#16x^2+5y^2+32x-10y-59=0#

#16(x^2+2x)+5(y^2-2y)=59#

#16(x^2+2x+1)+5(y^2-2y+1)=59+16+5#

#16(x+1)^2+5(y-1)^2=80#

#16(x+1)^2/80+5(y-1)^2/80=1#

#(x+1)^2/5+(y-1)^2/16=1#

#(x+1)^2/(sqrt5)^2+(y-1)^2/4^2=1#

This is an ellipse, center #=(-1,1)#

graph{(16x^2+32x+5y^2-10y-59)((x+1)^2+(y-1)^2-0.01)=0 [-11.25, 11.25, -5.625, 5.625]}

May 9, 2017

Answer:

Yes

Explanation:

To convert the equation to standard form for ellipses, we need to complete the square.
Note that the standard form for an ellipse is:
#((x-h)^2)/a^2+((y-k)^2)/b^2=1# with a horizontal semi-major axis
or
#((x-h)^2)/b^2+((y-k)^2)/a^2=1# with a vertical semi-major axis
where #(h,k)# is the center of the ellipse and #a>b#, where #a# is the length of the semi-major axis and #b# is the length of the semi-minor axis

Given the starting equation #16x^2+5y^2+32x-10y-59=0#, we first add 59 to both sides to isolate the variables and the constants:
#16x^2+5y^2+32x-10y=59#

Now we can complete the square for x and y:
#16(x^2+2x)+5(y^2-2y)=59#
#16(x+1)^2-16+5(y-1)^2-5=59#
#16(x+1)^2+5(y-1)^2=80#

Now we divide both sides by 80 to make the right-hand side equal to 1 for standard conic form:
#((x+1)^2)/5+((y-1)^2)/16=1#

Therefore, this conic is a vertical ellipse centered at #(-1,1)# with a semi-major axis of length #4# and a semi-minor axis of length #sqrt(5)#.