The eccentricity of the hyperbola (x2/a2)-(y2/b2)=1 which passes through the points (3,0) and (3 √2, 2), is

Question

The eccentricity of the hyperbola (x2/a2)-(y2/b2)=1 which passes through the points (3,0) and (3 √2, 2), is (A) (13/3) (B) √(13/3) (C) (√13/9) (D) (√13/3)

Tardigrade Admin · Accepted Answer

Given that the hyperbola (x2/a2)-(y2/b2)-1 is passing through the points (3,0) and (3 √2, 2), so we get a2=9 and b2=4. Again, we know that b2=a2(e2-1). This gives 4 =9(e2-1) or e2 =(13/9) or e =(√13/3)

Q. The eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ which passes through the points $(3, 0)$ and $(32, 2)$ , is

$\frac{13}{3}$

$\frac{13}{3}$

$\frac{13}{9}$

$\frac{13}{3}$

Given that the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} - 1$ is passing through the points $(3, 0)$ and $(32, 2)$ , so we get $a^{2} = 9$ and $b^{2} = 4$ .
Again, we know that $b^{2} = a^{2} (e^{2} - 1)$ . This gives
$4 = 9 (e^{2} - 1)$
or $e^{2} = \frac{13}{9}$
or $e = \frac{13}{3}$

Q. The eccentricity of the hyperbola a2x2​−b2y2​=1 which passes through the points (3,0) and (32​,2), is

313​

313​​

913​​

313​​

Given that the hyperbola a2x2​−b2y2​−1 is passing through the points (3,0) and (32​,2), so we get a2=9 and b2=4. Again, we know that b2=a2(e2−1). This gives 4=9(e2−1) or e2=913​ or e=313​​

Q. The eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ which passes through the points $(3, 0)$ and $(32, 2)$ , is

$\frac{13}{3}$

$\frac{13}{3}$

$\frac{13}{9}$

$\frac{13}{3}$