Question

The foci of the ellipse \[ \frac{x^2}{25}+\frac{y^2}{16}=1 \] and that of the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \] are same. The greatest length of the transverse axis of the hyperbola such that the difference of the squares of their eccentricities is at least one is

• A. \(\dfrac{15\sqrt2}{\sqrt{17}}\)
• B. \(6\)
• C. \(\dfrac{30}{\sqrt{17}}\)
• D. \(\dfrac{15}{\sqrt{34}}\)

Hint:

If an ellipse and a hyperbola have common foci, then \[ \boxed{a^2+b^2=c^2} \] for the hyperbola, where \(c\) is the common focal distance. Use \[ \boxed{e_{\text{ellipse}}^2=\frac{c^2}{a^2},\qquad e_{\text{hyperbola}}^2=\frac{c^2}{a^2}} \] to form the required inequality.

Answer 1

The Correct Option is A

Step 1: Find the common focal distance. For the ellipse, \[ a_1=5,\qquad b_1=4. \] Hence, \[ c^2=a_1^2-b_1^2 =25-16 =9, \] so \[ c=3. \] Since the foci are common, for the hyperbola \[ a^2+b^2=9. \]

Step 2: Use the condition on eccentricities. The eccentricity of the ellipse is \[ e_1^2=\frac{9}{25}. \] For the hyperbola, \[ e_2^2=\frac{c^2}{a^2} =\frac9{a^2}. \] Given, \[ e_2^2-e_1^2\ge1. \] Therefore, \[ \frac9{a^2}-\frac9{25}\ge1. \] Hence, \[ \frac9{a^2}\ge\frac{34}{25}, \] \[ a^2\le\frac{225}{34}. \]

Step 3: Find the greatest transverse axis. The greatest value of \[ a \] is \[ a=\frac{15}{\sqrt{34}}. \] Therefore, the greatest length of the transverse axis is \[ 2a = \frac{30}{\sqrt{34}} = \frac{15\sqrt2}{\sqrt{17}}. \] Hence, \[ \boxed{\frac{15\sqrt2}{\sqrt{17}}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.