Q.Equation of the hyperbola with eccentricity $\dfrac{3}{2}$ and foci at $(\pm 2, 0)$ is
(A) $\dfrac{x^2}{4} - \dfrac{y^2}{5} = \dfrac{4}{9}$
(B) $\dfrac{x^2}{9} - \dfrac{y^2}{9} = \dfrac{4}{9}$
(C) $\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1$
(D) none of these

The key idea is that for a hyperbola centred at the origin with foci on the x‑axis, the standard form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, with eccentricity $e = \frac{c}{a}$ and $c^2 = a^2 + b^2$.

Step 1: Foci are at $(\pm 2, 0)$, so $c = 2$. Given $e = \frac{3}{2}$, we have $e = \frac{c}{a} \implies \frac{3}{2} = \frac{2}{a} \implies a = \frac{4}{3}$.

Step 2: Using $c^2 = a^2 + b^2$: $4 = \frac{16}{9} + b^2 \implies b^2 = 4 - \frac{16}{9} = \frac{20}{9}$. …

The standard form of a hyperbola with centre at the origin and transverse axis along the x‑axis is

where $a$ is the distance from the centre to each vertex, and the foci are at $(\pm c, 0)$ with $c = ae$. The eccentricity $e = c/a$ is always greater than 1 for a hyperbola. The relation between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$, which can also be written as $b^2 = a^2(e^2 - 1)$.

Here the foci are given as $(\pm 2, 0)$, so $c = 2$. The eccentricity is $e = 3/2$. Since $c = ae$, we can find $a$ directly. Then $b^2$ follows from the relation above. Once $a^2$ and $b^2$ are known, we write the equation and compare with the options.

1. Find $a$ from $c = ae$. $c = 2$, $e = 3/2$, so

Hence $a^2 = \frac{16}{9}$.

2. Find $b^2$ using $b^2 = a^2(e^2 - 1)$. $e^2 = \frac{9}{4}$, so

Therefore

3. Write the standard equation. Substituting $a^2$ and $b^2$ into $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ gives

Multiply numerator and denominator to simplify:

4. Compare with the given options. Option (A) is $\frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}$. Multiply both sides of our equation by something to see if it matches. …