(ellipse)Reduce the equation 32𝑥2 + 50𝑦2 − 128𝑥 − 672 = 0 to standard form.

Question

Reduce the equation 32𝑥2 + 50𝑦2 − 128𝑥 − 672 = 0 to
standard form. Locate the center, foci, vertices, ends of latera
recta, and trace the curve.

Answer 1

32𝑥² + 50𝑦² − 128𝑥 − 672 = 0

 

To put into standard form, complete the square:

32(x² -4x + _____) +50 y² = 672.

 

To complete the square in the"x department" you will place the number 4 where I have the underlining.

DO YOU KNOW WHY?

 

The effect of putting 4 there is the net increase of the left side of this equation by 32(4), or 128, so to keep the equation balanced (true) we must add 128 to the constant on the right side. now we have

32(x2 -4x + 4) +50 y2 = 672 + 128  or  32(x2 -4x + 4) +50 y² = 800.

 

Now divide by 800, to make the constant on right side 1.

32(x - 2)² +50 y² = 1.

     800

 

This simplifies to (x-2)² +y² = 1.

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