Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
For the hyperbola $9y^2 - 16y^2 - 18x + 32y - 151 = 0$
one of the directrix is $x = 21/5$
length of latus rectum = 9/2
foci are (6, 1) and (-4, 1)
eccentricity is 5/4
Step-by-Step Solution
Key Concept: Translation of axes simplifies a conic to standard form, making it easy to identify parameters and compute eccentricity.
The hyperbola $\frac{(x-1)^2}{16}-\frac{(y-1)^2}{9}=1$ can be transformed using $X=x-1$ and $Y=y-1$ to standard form $\frac{X^2}{16}-\frac{Y^2}{9}=1$. With $a^2=16$ and $b^2=9$, the eccentricity is $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$.
Correct Answer: 1,2,3,4