Question

If the foci of the ellipse \[ \frac{x^2}{25}+\frac{y^2}{k^2}=1 \qquad (k^2<25) \] and the hyperbola \[ \frac{x^2}{k}-\frac{y^2}{5}=1 \] are the same, then the product of the lengths of the latus rectum of the ellipse and that of the hyperbola is:

• A. \(25\)
• B. \(50\)
• C. \(16\)
• D. \(32\)

Hint:

Remember the latus rectum formulas: For an ellipse: \[ L=\frac{2b^2}{a} \] For a hyperbola: \[ L=\frac{2b^2}{a} \] The formula is the same; only the values of \(a\) and \(b\) differ according to the conic.

Answer 1

The Correct Option is D

Concept: The foci of both conics are given to be identical. Therefore, their focal distances from the centre must be equal. For an ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the focal distance is \[ c^2=a^2-b^2. \] For a hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the focal distance is \[ c^2=a^2+b^2. \] Using the equality of the foci, we determine \(k\), then calculate the lengths of the latus recta.

Step 1: Find the focal distance of the ellipse. For \[ \frac{x^2}{25}+\frac{y^2}{k^2}=1, \] we have \[ a^2=25, \qquad b^2=k^2. \] Hence \[ c_e^2=25-k^2. \]

Step 2: Find the focal distance of the hyperbola. For \[ \frac{x^2}{k}-\frac{y^2}{5}=1, \] we have \[ a^2=k, \qquad b^2=5. \] Therefore \[ c_h^2=k+5. \]

Step 3: Use the fact that the foci are the same. Thus \[ c_e=c_h. \] Hence \[ 25-k^2=k+5. \] \[ k^2+k-20=0. \] \[ (k+5)(k-4)=0. \] Since \(k>0\), \[ k=4. \]

Step 4: Length of latus rectum of the ellipse. For an ellipse, \[ L_e=\frac{2b^2}{a}. \] Substituting \[ a=5,\qquad b^2=16, \] \[ L_e=\frac{2(16)}5 =\frac{32}{5}. \]

Step 5: Length of latus rectum of the hyperbola. For a hyperbola, \[ L_h=\frac{2b^2}{a}. \] Here \[ a^2=4 \Rightarrow a=2, \] and \[ b^2=5. \] Therefore, \[ L_h=\frac{2(5)}2 =5. \]

Step 6: Compute the required product. \[ L_eL_h = \frac{32}{5}\times5. \] \[ =32. \]

Step 7: Final Answer. \[ \boxed{32} \]