Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $16x^2 - 9y^2 = 576$.

Concept: Hyperbola Standard Form — we first rewrite the given equation in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Step 1: Divide both sides by 576:

Thus $a^2 = 36$, $b^2 = 64$, so $a = 6$, $b = 8$.

Step 2: For a hyperbola of this form, $c^2 = a^2 + b^2 = 36 + 64 = 100$, so $c = 10$.

Step 3: Now read off the required values:

• Vertices: $(\pm a, 0) = (\pm 6, 0)$
• Foci: $(\pm c, 0) = (\pm 10, 0)$ …

The given equation is $16x^2 - 9y^2 = 576$. This is a hyperbola, and the first step is always to rewrite it in the standard form so we can read off the key parameters.

Why standard form? The standard forms for a hyperbola centered at the origin are:

• Horizontal transverse axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
• Vertical transverse axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Here, the $x^2$ term is positive and the $y^2$ term is negative, so it's the horizontal case. The numbers $a$ and $b$ give us the vertices, and $c = \sqrt{a^2 + b^2}$ gives the foci. Eccentricity $e = c/a$, and latus rectum length is $2b^2/a$ for a horizontal hyperbola.

Let's work through it.

1. Divide through by 576 to get the right-hand side equal to 1.

Simplify each fraction:

2. Identify $a^2$ and $b^2$. From $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have $a^2 = 36$ and $b^2 = 64$. So:

3. Find $c$. For a hyperbola, $c^2 = a^2 + b^2$ (not $a^2 - b^2$ — that's for ellipses). So:

4. Vertices. For a horizontal hyperbola, vertices are at $(\pm a, 0) = (\pm 6, 0)$.

5. Foci. Foci are at $(\pm c, 0) = (\pm 10, 0)$. …