Mathematics | VITEEE 2018 | Discussion Page

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The equation of the hyperbola with vertices (3 , 0), (-3 , 0) and semi-latus rectum 4 is given by :

$4x^{2}-3y^{2}+36=0$

$4x^{2}-3y^{2}+12=0$

$4x^{2}-3y^{2}-36=0$

$4x^{2}+3y^{2}-25=0$

Option C

We have a = 3 and $\frac{b^{2}}{a}$ = 4

→ b² = 12

Hence, the equation of the hyperbola is

$\frac{x^{2}}{9}-\frac{y^{2}}{12}=1$

$4x^{2}-3y^{2}= 36$ = $4x^{2}-3y^{2}-36=0$

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