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 :

Hyperbola PYQ — JEE Main 2025 · 22 Jan · Shift 1 · Q26 with Solution

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Correct answer: C

The foci are given as $(1,14)$ and $(1,-12)$ .

Since the $x$ -coordinates are same, the hyperbola has a vertical transverse axis.

The center is the midpoint of the foci:

\left(\frac{1+1}{2},\frac{14+(-12)}{2}\right)=(1,1)

So the center is $(1,1)$ .

For a vertical hyperbola, the foci are $(h,k\pm c)$ .

Here,

c=14-1=13

So,

c=13

Since the transverse axis is vertical, the standard form is

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

Thus,

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

Also, for a hyperbola,

c^2=a^2+b^2

Hence,

169=a^2+b^2

Since the hyperbola passes through $(1,6)$ , substitute it:

\frac{(6-1)^2}{a^2}-\frac{(1-1)^2}{b^2}=1

\frac{25}{a^2}=1

So,

a^2=25

Using

a^2+b^2=c^2

we get

25+b^2=169

b^2=144

For a hyperbola, the length of the latus rectum is

\frac{2b^2}{a}

Here,

a=5,\quad b^2=144

Therefore,

\text{Length of latus rectum}=\frac{2\cdot 144}{5}=\frac{288}{5}

\frac{288}{5}

corresponds to Option C.