**20.** Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is: - (1) \( \frac{25}{6} \) - (2) \( \frac{24}{5} \) - (3) \( \frac{288}{5} \) - (4) \( \frac{144}{5} \)

1. Recognising the Hyperbola Type

Because both foci share the same $x$‐coordinate ($x=1$), they sit on a vertical line.
Hence the transverse axis (the axis that actually contains both foci) is vertical.
For such a hyperbola, the standard form is

where $(h,k)$ is the centre, $a$ measures half the length of the transverse axis, and $b$ measures half the length of the conjugate axis.

2. Finding the Centre $(h,k)$

The centre is the mid‐point of the two foci:

3. Getting $c$

Distance from the centre to either focus is $c$:

4. Hooking in the Given Point

Point $(1,6)$ must satisfy the equation. Plugging $x=1$ (hence $x-h=0$) and $y=6$ (so $y-k = 5$) into the general form gives

5. Relating $a$, $b$, and $c$

For any hyperbola,

6. Length of the Latus Rectum

For a hyperbola, the (full) latus‐rectum length $\ell$ is

Substituting $b^2 = 144$ and $a = 5$:

That matches option (3).

🚀 What’s Happening?

Imagine you have two nails fixed on a board at points F₁ = (1,14) and F₂ = (1,−12).
A hyperbola is a special loop‐shaped curve such that for every point on it, the difference of its distances to those two nails stays constant.

Now, you’re told the curve passes through P = (1,6) (one of its own points).
Your job: figure out how long a particular small line segment on this curve is, called the latus rectum.
Think of the latus rectum as a tiny “ladder” attached to the curve near a focus—its length depends completely on how wide and tall the hyperbola is.

So, in plain words:

1. Find the centre of the hyperbola (mid‐point of the two foci).
2. Work out the numbers that make its equation.
3. Insert the given point to pin those numbers down.
4. Use a ready‐made formula to spit out the latus‐rectum length. That’s it! ✨

Step-by-Step Solution

1. Centre $(h,k)$

   $$h = 1, \quad k = \frac{14 + (-12)}{2} = 1.$$

   So the standard form is

   $$\frac{(y-1)^2}{a^2} - \frac{(x-1)^2}{b^2} = 1.$$

2. Distance $c$ (focus to centre)

   $$c = |14 - 1| = 13 \quad \Rightarrow \quad c^2 = 169.$$

3. Use the point $(1,6)$

   $$\frac{(6-1)^2}{a^2} = 1 \quad \Rightarrow \quad \frac{25}{a^2} = 1 \quad \Rightarrow \quad a^2 = 25,\; a = 5.$$

4. Compute $b^2$

   $$c^2 = a^2 + b^2 \quad \Rightarrow \quad 169 = 25 + b^2 \quad \Rightarrow \quad b^2 = 144.$$

5. Length of latus rectum ($\ell$)

   $$\ell = \frac{2b^2}{a} = \frac{2 \times 144}{5} = \frac{288}{5}.$$

6. Choose the correct option
   Option (3) $\frac{288}{5}$.