Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $49y^2 - 16x^2 = 784$.

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

Step 1: Divide both sides by 784:

Step 2: Identify $a^2 = 16$, $b^2 = 49$, so $a = 4$, $b = 7$. For a hyperbola of this form, $c^2 = a^2 + b^2 = 16 + 49 = 65$, hence $c = \sqrt{65}$.

Step 3:

• Vertices: $(0, \pm a) = (0, \pm 4)$
• Foci: $(0, \pm c) = (0, \pm \sqrt{65})$ …

Concept First — Why This Approach Works

A hyperbola is defined by the difference of distances from two fixed points (foci) being constant. The standard form tells us everything about its shape and orientation at a glance.

For a vertical hyperbola (opening up and down), the standard form is:

Here $a$ is the distance from the centre to each vertex (along the y-axis), $b$ relates to the asymptotes, and $c = \sqrt{a^2 + b^2}$ gives the distance from the centre to each focus. The eccentricity $e = c/a$ tells how "stretched" the hyperbola is, and the latus rectum is a chord through a focus parallel to the conjugate axis.

The key insight: the positive term tells you which axis the hyperbola opens along. If $y^2$ is positive, it's vertical; if $x^2$ is positive, it's horizontal. This single observation prevents half the mistakes students make.

Step-by-Step Solution

1. Rewrite the equation in standard form.

We start with $49y^2 - 16x^2 = 784$. Divide every term by 784:

Simplify each fraction:

Now it's in the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 49$.

2. Identify $a$, $b$, and compute $c$.

From $a^2 = 16$, we get $a = 4$ (distance from centre to each vertex).

From $b^2 = 49$, we get $b = 7$ (related to asymptote slopes).

For any hyperbola, $c^2 = a^2 + b^2$, so:

3. Find the vertices.

For a vertical hyperbola centred at $(0,0)$, the vertices are at $(0, \pm a)$. So:

4. Find the foci.

The foci lie on the same axis as the vertices, further from the centre. For a vertical hyperbola, they are at $(0, \pm c)$: …