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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

Best Answer 

 #4
avatar+91411 
+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
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5+0 Answers

 #1
avatar+91411 
+10
Melody  Mar 27, 2015
 #5
avatar+776 
0

OMG!!! you use desmos to!!!!!!!

User101  Apr 16, 2016
 #2
avatar+91411 
+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

http://www.mathsisfun.com/geometry/ellipse.html

http://en.wikipedia.org/wiki/Ellipse

Melody  Mar 27, 2015
 #3
avatar+91411 
+3

Oh and Welcome to the forum Tyler :)

Melody  Mar 27, 2015
 #4
avatar+91411 
+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

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