Question

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) $\dfrac{\pi }{2}$

Answer 1

Hint: 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.