 Let a and b respectively be the semi - transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9{e^2} - 18e + 5 = 0$. If $S\left( {5,0} \right)$ is a focus and $5x = 9$ is the corresponding directrix of this hyperbola, then find the value of ${a^2} - {b^2}$ (A) $7$ (B) $5$ (C) $- 5$ (D) $- 7$ Verified
147.3k+ views
Hint:Firstly, start with finding the value of eccentricity from the quadratic equation. Now equate the focus point with $\left( {ae,0} \right)$ to find $a$. Now use the equation $e = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}}$ to find the value of $b$.

Let’s first look at all the information given in the question. We have here a quadratic equation $9{e^2} - 18e + 5 = 0$ that satisfies eccentricity. Then we have the focus point as $S\left( {5,0} \right)$ and equation of directrix $5x = 9$.
In a simple sense, hyperbola looks similar to mirrored parabolas. The two halves are called the branches. When the planes intersect on the halves of a right circular cone angle of which will be parallel to the axis of the cone, a parabola is formed.
A hyperbola contains two foci and two vertices. The foci of the hyperbola are away from the hyperbola’s centre and vertices.
According to the question, the equation $9{e^2} - 18e + 5 = 0$ has two solutions of eccentricity
$\Rightarrow 9{e^2} - 18e + 5 = 0 \Rightarrow 9{e^2} - 15e - 3e + 5 = 0 \Rightarrow \left( {3e - 1} \right)\left( {3e - 5} \right) = 0 \Rightarrow e = \dfrac{5}{3},\dfrac{1}{3}$
So, we got eccentricity value as $e = \dfrac{5}{3},\dfrac{1}{3}$. But we know that the eccentricity of a hyperbola cannot be less than one.So we take $\Rightarrow e = \dfrac{5}{3}$
Also, the focus coordinates for a hyperbola is $\left( {ae,0} \right)$ , but we already know that focus is at $S\left( {5,0} \right)$
$\Rightarrow ae = 5 \Rightarrow a = \dfrac{5}{e}$
After using the value$e = \dfrac{5}{3}$, we get: $\Rightarrow a = \dfrac{5}{e} = \dfrac{5}{{\dfrac{5}{3}}} = 3$
Also, the eccentricity of a hyperbola can be written as: $e = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}}$
Now, substituting all the known value in it, we get:
$\Rightarrow {\left( {\dfrac{5}{3}} \right)^2} = \dfrac{{{3^2} + {b^2}}}{{{3^2}}} \Rightarrow {b^2} = \dfrac{{{5^2}}}{{{3^2}}} \times {3^2} - {3^2} \Rightarrow b = \sqrt {25 - 9} = 4$
Therefore, we have the value of semi - transverse and semi-conjugate axes of a hyperbola as $a = 3$ and $b = 4$
Hence, the value of ${a^2} - {b^2} = {3^2} - {4^2} = 9 - 16 = - 7$

So, the correct answer is “Option D”.

Note:Try to figure out all the unknowns that go need for evaluating the given expression. Always keep in mind that the eccentricity of the hyperbola is greater than one. The line that passes through the centre, the focus of the hyperbola and vertices is the Major Axis. The line perpendicular to the major axis and passes by the middle of the hyperbola are the Minor Axis.