QUESTION

# The length of the transverse axis of a hyperbola is $2\cos \alpha$. The foci of the hyperbola are the same as that of the ellipse $9{x^2} + 16{y^2} = 144$ , the equation of the hyperbola isA. $\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 - {{\cos }^2}\alpha }} = 1$ B. $\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 + {{\cos }^2}\alpha }} = 1$ C. $\dfrac{{{x^2}}}{{1 + {{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 - {{\cos }^2}\alpha }} = 1$ D. $\dfrac{{{x^2}}}{{1 + {{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 + {{\cos }^2}\alpha }} = 1$ E. $\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{5 - {{\cos }^2}\alpha }} = 1$

Hint: First take the equation of hyperbola and solve it step by step adding the values and then substitute the value of a and b in the hyperbola equation and compare with the given option.

Let the equation of the hyperbola be $\dfrac{{{x^2}}}{{a_1^2}} - \dfrac{{{y^2}}}{{b_1^2}} = 1$
We know that the length of the transverse axis of hyperbola is $2{a_1}$ .
In the question it is given that the length of the transverse axis of hyperbola is $2\cos \alpha$.
$\Rightarrow 2{a_1} = 2\cos \alpha \\ \Rightarrow {a_1} = \cos \alpha \\$
We know that the focus of hyperbola is ${a_1}{e_1}$ and the focus of ellipse is $ae$.
Then equation of ellipse is $9{x^2} + 16{y^2} = 144$
$\Rightarrow \dfrac{{9{x^2}}}{{144}} + \dfrac{{16{y^2}}}{{144}} = 1 \\ \Rightarrow \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1 \\$
General equation of ellipse with origin as centre is,
$\Rightarrow \dfrac{{{x^2}}}{{a^2}} + \dfrac{{{y^2}}}{b^2} = 1 \\$
Comparing the above equations, we get, ${a^2} = 16$ and ${b^2} = 9$.
$\therefore a = 4$ and $b = 3$ (as a & b denote lengths, they can’t take negative values)
Now we have to find the eccentricity of ellipse, we know that the formula of eccentricity of ellipse is $\sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}}$
By putting the value of a and b
$e = \sqrt {1 - \dfrac{9}{{16}}} = \dfrac{{\sqrt 7 }}{4}$
In the question it is given that the foci of the hyperbola and ellipse are equal so ${a_1}{e_1} = ae$
$\Rightarrow {e_1}\cos \alpha = 4\dfrac{{\sqrt 7 }}{4}\because \left( {{a_1} = \cos \alpha } \right) \\ \Rightarrow {e_1}\cos \alpha = \sqrt 7 \\$
We know that eccentricity of hyperbola, ${e_1} = \sqrt {1 + \dfrac{{b_1^2}}{{a_1^2}}}$
Substituting, we get,
$\Rightarrow \cos \alpha \sqrt {1 + \dfrac{{b_1^2}}{{{{\cos }^2}\alpha }}} = \sqrt 7 \\ \Rightarrow \dfrac{{\cos \alpha }}{{\cos \alpha }}\sqrt {{{\cos }^2}\alpha + b_1^2} = \sqrt 7 \\$
Squaring both the sides, we get
$\Rightarrow {\cos ^2}\alpha + b_1^2 = 7 \\ \Rightarrow b_1^2 = 7 - {\cos ^2}\alpha \\$
Now put the value of ${a_1}$ and ${b_1}$ in equation of hyperbola,
$\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 - {{\cos }^2}\alpha }} = 1$
Therefore the required solution of hyperbola is $\dfrac{{{x^2}}}{{{{\cos }^2}\alpha }} - \dfrac{{{y^2}}}{{7 - {{\cos }^2}\alpha }} = 1$, so the correct answer is option A.

Note: In this type of equation first assume the equation of hyperbola and ellipse, then using the necessary formulas of eccentricities and the given conditions in the problem statements, determine the values of the constants.