Q27
jee_main_2024_31_jan_morning
Ellipse and Hyperbola Properties
Let the foci and length of the latus rectum of an ellipse fracx^2a^2 + fracy^2b^2 = 1, a > b be (pm 5, 0) and sqrt50, respectively. Then, the square of the eccentricity of the hyperbola fracx^2b^2 - fracy^2a^2 b^2 = 1 equals
Numerical Answer. Answer: 51 to 51
Solution
### Core Logic
For the ellipse, foci are at (pm 5, 0) implies ae = 5.
Latus rectum = frac2b^2a = sqrt50 = 5sqrt2 implies b^2 = frac5sqrt2a2.
### Step 1: Solve for a and b
Using b^2 = a^2(1 - e^2):
a^2 - (ae)^2 = b^2 implies a^2 - 25 = frac5sqrt2a2
2a^2 - 5sqrt2a - 50 = 0
2a^2 - 10sqrt2a + 5sqrt2a - 50 = 0
2a(a - 5sqrt2) + 5sqrt2(a - 5sqrt2) = 0
a = 5sqrt2 (since a > 0).
Now, b^2 = frac5sqrt2(5sqrt2)2 = 25 implies b = 5.
### Step 2: Hyperbola Eccentricity
The hyperbola is fracx^2b^2 - fracy^2a^2b^2 = 1.
Here, semi-major axis A = b and semi-minor axis B = ab.
Using eccentricity formula for hyperbola e_H^2 = 1 + fracB^2A^2:
e_H^2 = 1 + fraca^2 b^2b^2 = 1 + a^2
Since a = 5sqrt2, a^2 = 50.
e_H^2 = 1 + 50 = 51
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Conic Sections
