What are the conic sections of the following equations #16x^2 + 25y^2- 18x - 20y + 8=0#?

1 Answer
May 16, 2018

Answer:

It is an ellipse.

Explanation:

The above equation can be easily converted into the ellipse form #(x-h)^2/a^2+(y-k)^2/b^2=1# as coefficients of #x^2# and#y^2# both are positive), where #(h,k)# is the center of ellipse and axis are #2a# and #2b#, with larger one as major axis an other minor axis. We can also find vertices by adding #+-a# to #h# (keeping ordinate same) and #+-b# to #k# (keeping abscissa same).

We can write the equation #16x^2+25y^2-18x-20y+8=0# as

#16(x^2-18/16x)+25(y^2-20/25y)=-8#

or #16(x^2-2*9/16x+(9/16)^2)+25(y^2-2*2/5y+(2/5)^2)=-8+16(9/16)^2+25(2/5)^2#

or #16(x-9/16)^2+25(y-2/5)^2=-8+81/16+4#

or #16(x-9/16)^2+25(y-2/5)^2=17/16#

or #(x-9/16)^2/(sqrt17/16)^2+(y-2/5)^2/(sqrt17/20)^2=1#

Hence center of ellipse is #(9/16,2/5)#, while major axis parallel to #x#-axis is #sqrt17/8# and minor axis parallel to #y#-axis is #sqrt17/10#.

graph{(16x^2+25y^2-18x-20y+8)((x-9/16)^2+(y-2/5)^2-0.0001)(x-9/16)(y-2/5)=0 [-0.0684, 1.1816, 0.085, 0.71]}