Comprehensive Step-by-Step Solution
Step 1: Evaluation of the Ellipse Parameters
The standard equation of the given ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$.
Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
$a^2 = 36 \implies a = 6$
$b^2 = 16 \implies b = 4$
The eccentricity of this ellipse ($e_e$) is:
$$e_e = \sqrt{1 – \frac{b^2}{a^2}} = \sqrt{1 – \frac{16}{36}} = \sqrt{\frac{20}{36}} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$$
Step 2: Locating the Foci
The foci of the ellipse are located at $(\pm ae_e, 0)$.
Distance of focus from origin ($c$) = $ae_e = 6 \times \frac{\sqrt{5}}{3} = 2\sqrt{5}$.
Therefore, the foci are $(\pm 2\sqrt{5}, 0)$.
Step 3: Deriving Hyperbola Dimensions
For the hyperbola, the foci are the same: $(\pm AE, 0) = (\pm 2\sqrt{5}, 0)$.
Thus, $AE = 2\sqrt{5}$.
Given the eccentricity of the hyperbola $E = 5$.
Semi-transverse axis $A = \frac{2\sqrt{5}}{E} = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$.
Now, find $B^2$ using $B^2 = A^2(E^2 – 1)$:
$$B^2 = \left(\frac{2}{\sqrt{5}}\right)^2 (5^2 – 1) = \frac{4}{5} \times 24 = \frac{96}{5}$$
Step 4: Final Calculation of Latus Rectum
Length of Latus Rectum = $\frac{2B^2}{A}$
$$L.R. = \frac{2 \times \frac{96}{5}}{\frac{2}{\sqrt{5}}} = \frac{96}{5} \times \sqrt{5} = \frac{96}{\sqrt{5}}$$
Deep-Dive Theory: Conic Sections and Focal Properties
1. The Concept of Confocal Conics
Confocal conics are curves that share the same focal points. This problem highlights a fundamental relationship in coordinate geometry where an ellipse and a hyperbola share their foci. Such curves are always orthogonal to each other at their points of intersection. In a system of confocal conics, every point in the plane (except the foci) lies on exactly one ellipse and one hyperbola from the family.
2. Ellipse Dynamics
An ellipse is defined by the constant sum of distances to two foci. The eccentricity $e$ measures the “flatness.” As $b \to a$, $e \to 0$ (circle). As $b \to 0$, $e \to 1$ (line segment). The focal distance $ae$ is the most critical parameter when bridging data to other conics.
3. Hyperbola Characteristics
Unlike the ellipse, the hyperbola is defined by the constant difference of focal distances. The eccentricity $E$ is always $>1$. The latus rectum $\frac{2B^2}{A}$ serves as a measure of how wide the hyperbola opens at the focus. For very high eccentricity (like $E=5$ in this case), the hyperbola’s branches become nearly parallel very quickly.
4. Advanced Geometry Application
The study of these parameters is vital for understanding trajectories. For example, in space mechanics, a satellite’s path might transition from an elliptical orbit to a hyperbolic escape trajectory if its velocity exceeds the escape velocity. The common focal point (the planet or star) remains the anchor for both mathematical models.
5. Calculation Shortcuts for JEE
In competitive exams, remember that if the foci are the same, then $(ae)_{ellipse} = (AE)_{hyperbola}$. Using this, you can directly relate the semi-axes if you know the eccentricities. Also, $B^2$ for a hyperbola can be written as $(AE)^2 – A^2$, which simplifies the substitution of focal distance.
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