If the eccentricity of the hyperbola \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\text{ }\!\!\alpha\!\!\text{ }=5\] is \[\sqrt{3}\] times the eccentricity of the ellipse \[{{x}^{2}}{{\sec }^{2}}\text{ }\!\!\alpha\!\!\text{ }+{{y}^{2}}=25\] , then the value of \[\alpha \] is
a) \[\dfrac{\pi }{6}\]
b) \[\dfrac{\pi }{4}\]
c) \[\dfrac{\pi }{3}\]
d) \[\dfrac{\pi }{2}\]
VerifiedHint: Since the give equation of ellipse is \[{{x}^{2}}{{\sec }^{2}}\alpha +{{y}^{2}}=25\] and hyperbola is \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\alpha =5\] .Convert the given equations into standard form of equation of ellipse and hyperbola:
Standard form of ellipse and hyperbola are given below:
Ellipse: \[\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1\]
Hyperbola:
\[ for\text{ }a\text{ }>\text{ }b\text{ }:\text{ }\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=\]
\[ for\text{ }a\text{ }<\text{ }b\text{ }:\text{ }\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}-\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}=1 \]
Then try to find the eccentricity for both the conic sections using the formula to find eccentricity. Eccentricity of both the conic sections is given below:
Ellipse:
\[ for\text{ }a\text{ }>\text{ }b\text{ }:\text{ }e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\text{ }\]
\[ for\text{ }a\text{ }<\text{ }b\text{ }:\text{ }e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}} \]
Hyperbola: \[e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}\]
After finding the eccentricity of both the equation in terms of \[\alpha \], equate the eccentricity of the hyperbola to \[\sqrt{3}\] times the eccentricity of the ellipse and find the value of \[\alpha \].
Complete step by step answer:
We have equation of ellipse as: \[{{x}^{2}}{{\sec }^{2}}\alpha +{{y}^{2}}=25\]as shown in the diagram above.
By dividing the above equation by 25, we can write it as:
\[\Rightarrow \dfrac{{{x}^{2}}}{5}-\dfrac{{{y}^{2}}}{5{{\cos }^{2}}\alpha }=1......(6)\]
Since ${{\sec }^{2}}\alpha =\dfrac{1}{{{\cos }^{2}}\alpha }$; we can write equation (1) as:
\[\Rightarrow \dfrac{{{x}^{2}}}{25{{\cos }^{2}}\alpha }+\dfrac{{{y}^{2}}}{25}=1......(2)\]
Where ${{a}^{2}}=25{{\cos }^{2}}\alpha $ and ${{b}^{2}}=25$
So, using the formula to find eccentricity of an ellipse: $e=\sqrt{1-\dfrac{{{a}^{2}}}{{{b}^{2}}}}$, the eccentricity of the ellipse \[{{x}^{2}}{{\sec }^{2}}\alpha +{{y}^{2}}=25\] can be written as:
\[ e=\sqrt{1-\dfrac{25{{\cos }^{2}}\alpha }{25}} \]
\[ =\sqrt{1-{{\cos }^{2}}\alpha }\]......(3)
Since ${{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1$
We can write equation (3) as:
\[ e=\sqrt{{{\sin }^{2}}\alpha } \]
\[ =\sin \alpha \]......(4)
Now, we have equation of hyperbola as: \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\alpha =5\] as shown in the figure above.
By dividing the above equation by 5, we can write it as:
\[\Rightarrow \dfrac{{{x}^{2}}}{5}-\dfrac{{{y}^{2}}{{\sec }^{2}}\alpha }{5}=1......(5)\]
Since${{\sec }^{2}}\alpha =\dfrac{1}{{{\cos }^{2}}\alpha }$; we can write equation (5) as:
\[\Rightarrow \dfrac{{{x}^{2}}}{5}-\dfrac{{{y}^{2}}}{5{{\cos }^{2}}\alpha }=1......(6)\]
Where ${{a}^{2}}=5$ and ${{b}^{2}}=5{{\cos }^{2}}\alpha $
So, using the formula to find eccentricity of hyperbola $e=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}}{a}$, the eccentricity of hyperbola \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\alpha =5\] can be written as:
$e=\sqrt{1+{{\cos }^{2}}\alpha }.....(7)$
As we know that, the eccentricity of the hyperbola is \[\sqrt{3}\] times the eccentricity of the ellipse.
So, we get a relation between equation (4) and equation (7) as:
$\sqrt{3}\times \sin \alpha =\sqrt{1+{{\cos }^{2}}\alpha }......(8)$
Square both the sides of equation (8), we get:
$3{{\sin }^{2}}\alpha =\left( 1+{{\cos }^{2}}\alpha \right)......(9)$
Since ${{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1$
We can write equation (9) as:
\[ \Rightarrow 3{{\sin }^{2}}\alpha =\left( 1+1-{{\sin }^{2}}\alpha \right) \]
\[ \Rightarrow 4{{\sin }^{2}}\alpha =2 \]
\[ \Rightarrow {{\sin }^{2}}\alpha =\dfrac{1}{2} \]
\[ \Rightarrow \sin \alpha =\dfrac{1}{\sqrt{2}} \]
Since $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Therefore, $\alpha =\dfrac{\pi }{4}$
So, the correct answer is “Option B”.
Note: It is easier to solve the equations regarding any conic section if they are simplified into their standard form. Otherwise, you may get a complicated equation as a result. So, always try to simplify the given equation of a conic section into its standard form.
Try to apply simplified trigonometric identities, else it might complicate the question.
