JEE Mathematics 2025 Question with Solution

Given: The foci are $(1,14)$ and $(1,-12)$, and the hyperbola passes through $(1,6)$. Find: The length of the latus-rectum.

The foci lie on the vertical line $x=1$, so the transverse axis is vertical. The centre is the midpoint of the foci:

$$(1,\tfrac{14+(-12)}{2})=(1,1)$$

Also, the distance between the foci is $26$, so

$$2c=26 \Rightarrow c=13$$

For a vertical hyperbola with centre $(1,1)$, the equation is

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

with

$$c^2=a^2+b^2$$

Since the point $(1,6)$ lies on the hyperbola, substituting gives

$$\frac{(6-1)^2}{a^2}-0=1$$ $$\frac{25}{a^2}=1 \Rightarrow a^2=25 \Rightarrow a=5$$

Now,

$$b^2=c^2-a^2=13^2-5^2=169-25=144$$

For the hyperbola $\dfrac{(y-1)^2}{a^2}-\dfrac{(x-1)^2}{b^2}=1$, the length of each latus-rectum is

$$\frac{2b^2}{a}$$

Therefore,

$$\text{length} = \frac{2\cdot 144}{5}=\frac{288}{5}$$

So, the length of the latus-rectum is $\frac{288}{5}$, and the correct option is C.

Given: The foci are $(1,14)$ and $(1,-12)$. Find: The length of the latus-rectum.

From the extracted working, the midpoint of the foci gives the centre $(1,1)$ and hence

$$c=13$$

The point $(1,6)$ is on the hyperbola, so the distance from the centre to this vertex-side point is

$$a=|6-1|=5$$

Thus,

$$a^2=25$$

Using

$$c^2=a^2+b^2$$

we get

$$b^2=13^2-5^2=169-25=144$$

Hence the latus-rectum length is

$$\frac{2b^2}{a}=\frac{2\cdot 144}{5}=\frac{288}{5}$$

The extracted first approach states $\frac{2a^2}{b}$, but that formula does not match the hyperbola obtained here. The correct computation from the full working gives $\frac{288}{5}$. Therefore, the correct option is C.