Identify the conic section.

4x^2 + 7y^2 + 32x - 56y + 148 = 0

Correct me if I'm wrong, but I think it's hyperbola with center (4, 4) and foci at (-4, 5.73), (-4, 2.27)?

thanks!

tylersomers2000
Mar 27, 2015

#4**+10 **

NOTE that your earlier hyperbola had a minus sign between the squared terms

and

The ellipse as a positive sign. If the numbers on the botton were the same then it would be a circle.

A circle is a special case of an elipse because it has a double focii instead of 2 separate focii.

Effectively a cirlce has one centre and other ellipses have 2 centres.

Melody
Mar 27, 2015

#2**+10 **

$$\\4x^2 + 7y^2 + 32x - 56y + 148 = 0 \\\\

4x^2+ 32x + 7y^2 - 56y =-148 \\\\

4(x^2+8x)+7(y^2-8y)=-148\\\\

4(x^2+8x+16)+7(y^2-8y+16)=-148+4*16+7*16\\\\

4(x+4)^2+7(y-4)^2=28\\\\

\frac{4(x+4)^2}{28}+\frac{7(y-4)^2}{28}=1\\\\

\frac{(x+4)^2}{7}+\frac{(y-4)^2}{4}=1\\\\

\frac{(x+4)^2}{(\sqrt7)^2}+\frac{(y-4)^2}{2^2}=1\\\\$$

This is an ellipse centre (-4,+4)

The ends of the major axis are $$(-4-\sqrt7,4) and (-4+\sqrt7,4)$$

Then ends of the minor axis are (-4,4-2) (-4,4+2) that is $$(-4,2) and (-4,6)$$

references

Melody
Mar 27, 2015

#4**+10 **

Best Answer

NOTE that your earlier hyperbola had a minus sign between the squared terms

and

The ellipse as a positive sign. If the numbers on the botton were the same then it would be a circle.

A circle is a special case of an elipse because it has a double focii instead of 2 separate focii.

Effectively a cirlce has one centre and other ellipses have 2 centres.

Melody
Mar 27, 2015