Question

Equation of the hyperbola with length of the latusrectum 4 and e = 3 is

• A. 2x²-16y² =1
• B. 3x²-15y² =1
• C. 16x²-2y² =1
• D. 8x²-y² =1

Answer 1

The correct option is C

16x²-2y² =1

$\frac{2b^2}{a}=4$ e = 3

$b^2=2a$

$a^2(e^2-1)=2a$

$\Rightarrow$ a(8) = 2 $\Rightarrow$ a = $\frac{1}{4}$ , $b^2=\frac{1}{2}$

$\frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{2}}=1$

$\Rightarrow$ 16 $x^2-2y^2=1$