Question

The foci of a hyperbola are the same as those of the ellipse \[ 9x^2+16y^2=144 \] If the length of the transverse axis of the hyperbola is $2\cos\alpha$, then its equation is:

• A. \[ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{7-\cos^2\alpha} =1 \]
• B. \[ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{7+\cos^2\alpha} =1 \]
• C. \[ \frac{x^2}{7-\cos^2\alpha} - \frac{y^2}{\cos^2\alpha} =1 \]
• D. \[ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{5-\cos^2\alpha} =1 \]

Hint:

For confocal conics centered at origin: \[ a^2-b^2=c^2 \] for ellipses and \[ A^2+B^2=c^2 \] for hyperbolas. Equating the same $c^2$ immediately connects both conics.

Answer 1

The Correct Option is A

Understanding the Concept: Two conics are said to be confocal if they share the same foci. For ellipse: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] the focal distance satisfies: \[ c^2=a^2-b^2 \] For hyperbola: \[ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1 \] the focal distance satisfies: \[ c^2=A^2+B^2 \]
Step 1: Convert the ellipse into standard form.
Given: \[ 9x^2+16y^2=144 \] Divide throughout by $144$: \[ \frac{x^2}{16}+\frac{y^2}{9}=1 \] Thus, \[ a^2=16,\qquad b^2=9 \]
Step 2: Find the focal distance of the ellipse.
For ellipse: \[ c^2=a^2-b^2 \] Therefore, \[ c^2=16-9 \] \[ c^2=7 \] Hence the common foci are: \[ (\pm\sqrt7,0) \]
Step 3: Find the transverse axis parameter of the hyperbola.
Length of transverse axis: \[ 2A=2\cos\alpha \] Thus, \[ A=\cos\alpha \] Hence, \[ A^2=\cos^2\alpha \]
Step 4: Determine $B^2$.
For hyperbola: \[ c^2=A^2+B^2 \] Substitute values: \[ 7=\cos^2\alpha+B^2 \] Therefore, \[ B^2=7-\cos^2\alpha \]
Step 5: Write the equation of the hyperbola.
Standard form: \[ \frac{x^2}{A^2}-\frac{y^2}{B^2}=1 \] Substituting values: \[ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{7-\cos^2\alpha} =1 \] Hence, \[ \boxed{ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{7-\cos^2\alpha} =1 } \]