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) $\frac{144}{5}$
Solution:
The foci are $S_1(1, 14)$ and $S_2(1, -12)$.
Since the x-coordinates are same, it is a vertical hyperbola.
Distance between foci $2be = |14 – (-12)| = 26 \Rightarrow be = 13$.
Centre of hyperbola = Midpoint of foci = $(\frac{1+1}{2}, \frac{14-12}{2}) = (1, 1)$.
The hyperbola passes through $(1, 6)$. Since this point lies on the axis of the hyperbola (line $x=1$), it must be a vertex.
Distance from Centre $(1, 1)$ to Vertex $(1, 6)$ is $b$.
$$ b = |6 – 1| = 5 $$
We have $b = 5$ and $be = 13$.
Relation for hyperbola: $a^2 = b^2(e^2 – 1) = (be)^2 – b^2$.
$$ a^2 = (13)^2 – (5)^2 = 169 – 25 = 144 $$
Length of Latus Rectum ($LR$) for vertical hyperbola $\frac{(y-k)^2}{b^2} – \frac{(x-h)^2}{a^2} = 1$:
$$ LR = \frac{2a^2}{b} $$
$$ LR = \frac{2(144)}{5} = \frac{288}{5} $$
Ans. (3)