Question

Does anyone know the minor axis vertices of 16x^2+4y^2+64 x+24y+36=0? The major axis vertices are (-2,-7),(-2,1), but what are the minor?

Answer 1

This is the equation of an ellipse translated (slide) away from (0,0). You can use "complete the square" to find the new origin. I'll do the x terms and leave the y for you:

\[16x ^{2}+64x = 16(x ^{2}+4x+c)-16c\]
I factored out the 16 to make the numbers smaller. We need to find c to complete the square. but c= ( coeff on x term/2) squared. i.e. c= (4/2)^2= 4. Of course we can't add in c*16 without adding -16c to keep things in balance. Your new equation is
\[16(x ^{2}+4x+4)-64 + 4y^{2}+24y+36= 0\]
\[16(x+2)^{2}-64+ 4y^{2}+24y+36= 0\]
Do the same for y, re-arrange the equation into the standard form for an ellipse
\[\frac{(x-x _{0})}{a^{2}} +\frac{(y-y _{0})}{b^{2}} =1\]
a is the semi-axes of x (the ellipse vertex on the x axis is x0+a,y0 where x0,y0 is the center of the ellipse. BTW, the semi-minor axis is the smaller of a and b. The semi-major axis is the larger of a and b.