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Question: If the foci of \(\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\) coincide with the foc...

If the foci of x2a2−y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 coincide with the foci of x225+y29=1\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1 and eccentricity of the hyperbola is 2 then:
(a)a2+b2=14{{a}^{2}}+{{b}^{2}}=14
(b)There is a director circle of the hyperbola
(c)Centre of the circle if (0, 0)
(d)Length of the latus rectum of the hyperbola is 12

Explanation

Solution

We have given the equation of ellipse whose centre is at (0, 0) is x225+y29=1\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1. From this equation we have a2=25{{a}^{2}}=25 and b2=9{{b}^{2}}=9 then we will find its eccentricity using the formula e=a2−b2a2e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}. Now, we know that foci of ellipses are equal to (±ae,0)\left( \pm ae,0 \right) then substitute the values of “a” and “e” in it. And as it is given that foci of ellipse and hyperbola and we know the value of “ae” which we have derived above then substitute the value of “e” as 2 in “ae” because, in the above problem, it is given that eccentricity of the hyperbola is 2. From this, we will get the value of “a” and “b” for hyperbola. Then we can solve the options given above very easily.

Complete step-by-step solution:
The equation of ellipse given above is as follows:
x225+y29=1\dfrac{{{x}^{2}}}{25}+\dfrac{{{y}^{2}}}{9}=1
This equation is written in the form of standard ellipse equation which is:
x2a2+y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1
So, the value of a2&b2{{a}^{2}}\And {{b}^{2}} of the given ellipse is 25 and 9 respectively. We know that eccentricity for ellipse is given by:
e=a2−b2a2e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}
Substituting a2=25{{a}^{2}}=25 and b2=9{{b}^{2}}=9 in the above equation we get,
e=25−925 ⇒e=1625 ⇒e=45 \begin{aligned} & e=\sqrt{\dfrac{25-9}{25}} \\\ & \Rightarrow e=\sqrt{\dfrac{16}{25}} \\\ & \Rightarrow e=\dfrac{4}{5} \\\ \end{aligned}
We also know that foci of the ellipse is equal to (±ae,0)\left( \pm ae,0 \right) so substituting the value of “e and a” in this foci we get,
(±5(45),0) =(±4,0) \begin{aligned} & \left( \pm 5\left( \dfrac{4}{5} \right),0 \right) \\\ & =\left( \pm 4,0 \right) \\\ \end{aligned}
Now, it is given that foci of hyperbola and ellipse are same so foci of hyperbola is equal to:
(±4,0)\left( \pm 4,0 \right)
This means that ae=4ae=4 in the hyperbola.
We have given the equation of hyperbola as x2a2−y2b2=1\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 and we have given the eccentricity of a hyperbola as 2 so substituting the value of “e” as 2 in ae=4ae=4 we get,
a(2)=4 ⇒a=42=2 \begin{aligned} & a\left( 2 \right)=4 \\\ & \Rightarrow a=\dfrac{4}{2}=2 \\\ \end{aligned}
We know the formula for eccentricity of hyperbola as:
b2=a2(e2−1){{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right)
Substituting “a” as 2 and “e” as 2 in the above equation we get,
b2=(2)2((2)2−1) ⇒b2=4(4−1) \begin{aligned} & {{b}^{2}}={{\left( 2 \right)}^{2}}\left( {{\left( 2 \right)}^{2}}-1 \right) \\\ & \Rightarrow {{b}^{2}}=4\left( 4-1 \right) \\\ \end{aligned}
⇒b2=4(3) ⇒b2=12 \begin{aligned} & \Rightarrow {{b}^{2}}=4\left( 3 \right) \\\ & \Rightarrow {{b}^{2}}=12 \\\ \end{aligned}
Now, addition of square of “a” and square of “b” is equal to:
a2+b2=(2)2+12 ⇒a2+b2=4+12=16 \begin{aligned} & {{a}^{2}}+{{b}^{2}}={{\left( 2 \right)}^{2}}+12 \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}=4+12=16 \\\ \end{aligned}
Hence, option (a) is incorrect.
From the above, you can see that b2>a2{{b}^{2}}>{{a}^{2}} so there is no director circle of the hyperbola.
So, option (b) and option (c) is automatically cancelled out.
We know that, length of latus rectum of the hyperbola is equal to:
2b2a\dfrac{2{{b}^{2}}}{a}
Substituting the value of a=2&b2=12a=2\And {{b}^{2}}=12 in the above expression we get,
2(12)2=12\dfrac{2\left( 12 \right)}{2}=12
Hence, option (d) is correct.

Note: The question demands the knowledge of eccentricity, foci of the ellipse, and hyperbola. Also, you should know the condition for the director circle to the hyperbola failing of any concept will paralyze you in going further in the above problem.
You might get confused in the formula for the eccentricity of ellipse and hyperbola. In the below, we are shown the different eccentricity formula for ellipse and hyperbola.
Eccentricity for ellipse:
e2=1−b2a2{{e}^{2}}=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}
Eccentricity for hyperbola:
e2=1+b2a2{{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}
Now, you can see from the above formula that there is only one significant difference in the formulae of ellipse and hyperbola so you can remember as the eccentricity for an ellipse is less than 1 and the eccentricity of the hyperbola is greater than 1.