Class 11

Math

3D Geometry

Conic Sections

Find the equation of the hyperbola satisfying the give conditions: Foci $(±4,0)$ the latus rectum is of length $12$

Therefore, the equation of the hyperbola is of the form $a_{2}x_{2} −b_{2}y_{2} =1$

Since the foci are $(±4,0)⇒ae=c=4$

Length of latus rectum $=12$

$⇒a2b_{2} =12$

$⇒$ $b_{2}$ $=6a$

We know that $a_{2}+b_{2}=c_{2}$

$∴a_{2}+6a=16$

$⇒a_{2}+6a−16=0$

$⇒a_{2}+8a−2a−16=0$

$⇒(a+8)(a−2)=0$

$⇒a=−8,2$

Since a is non-negative $a=2$

$∴b_{2}=6a=6×2=12$

Thus the equation of the hyperbola is $4x_{2} −12y_{2} =1$