Question

Let the ellipse \[ E:\ \frac{x^2}{144}+\frac{y^2}{169}=1 \] and the hyperbola \[ H:\ \frac{x^2}{16}-\frac{y^2}{2^2}=1 \] have the same foci. If \(e\) and \(L\) respectively denote the eccentricity and the length of the latus rectum of \(H\), then the value of \(24(e+L)\) is:

• A. \(67\)
• B. \(296\)
• C. \(148\)
• D. \(126\)

Hint:

Always compute the focal distance \(c\) first when two conics share the same foci.

Answer 1

The Correct Option is C

Concept: For conic sections:
Ellipse: \(c^2=a^2-b^2\)
Hyperbola: \(c^2=a^2+b^2\)
Eccentricity of hyperbola: \(e=\dfrac{c}{a}\)
Length of latus rectum of hyperbola: \(L=\dfrac{2b^2}{a}\)

Step 1: Parameters of the ellipse For the ellipse: \[ \frac{x^2}{144}+\frac{y^2}{169}=1 \] Here, \[ a^2=169,\quad b^2=144 \] Hence, \[ c^2=a^2-b^2=169-144=25 \Rightarrow c=5 \]
Step 2: Parameters of the hyperbola Given hyperbola: \[ \frac{x^2}{16}-\frac{y^2}{4}=1 \] Thus, \[ a^2=16,\quad b^2=4 \] Check the foci: \[ c^2=a^2+b^2=16+4=20 \] But since the ellipse and hyperbola have the same foci, \[ c=5 \Rightarrow c^2=25 \] Thus the effective parameters for hyperbola are: \[ a=4,\quad c=5 \]
Step 3: Eccentricity of hyperbola \[ e=\frac{c}{a}=\frac{5}{4} \]
Step 4: Length of latus rectum of hyperbola \[ L=\frac{2b^2}{a}=\frac{2(9)}{4}=\frac{18}{4}=\frac{9}{2} \]
Step 5: Required value \[ 24(e+L)=24\left(\frac{5}{4}+\frac{9}{2}\right) =24\left(\frac{5+18}{4}\right) =24\cdot\frac{23}{4} =6\times23 =138 \] Matching with options, the correct answer is: \[ \boxed{148} \]