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Question: Given that, \({E_1} = \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1 = 0,\) (a>b) and \({E_...

Given that, E1=x2a2+y2b2−1=0,{E_1} = \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1 = 0, (a>b) and E2=x2k2+y2b2−1=0,{E_2} = \dfrac{{{x^2}}}{{{k^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1 = 0, , (k < b) E2{E_2} is inscribed in E1{E_1} . If E1{E_1} and E2{E_2} have the same eccentricities then the length of minor axis of E2=p{E_2} = p (LLR of E1{E_1} ) then p =?

Explanation

Solution

In the above, you have to find the value of p for the given condition. Also, in the question it is mentioned the length of the minor axis, for an ellipse formula of the length of the minor axis is 2b. Note, that first, you will have to convert the equation into the standard form of an ellipse. So let us see how we can solve this problem.

Complete step by step answer:
Given that, E1=x2a2+y2b2−1=0{E_1} = \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1 = 0 and E2=x2k2+y2b2−1=0{E_2} = \dfrac{{{x^2}}}{{{k^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1 = 0 .
So, x2a2+y2b2=1\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 and x2k2+y2b2=1\dfrac{{{x^2}}}{{{k^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1
It is mentioned in the question that the eccentricities of both E1{E_1} and E2{E_2} are the same. So,
⇒b2a2+1=k2b2+1\Rightarrow \dfrac{{{b^2}}}{{{a^2}}} + 1 = \dfrac{{{k^2}}}{{{b^2}}} + 1
Subtracting 1 on each side of the above equation, we get
⇒b2a2=k2b2\Rightarrow \dfrac{{{b^2}}}{{{a^2}}} = \dfrac{{{k^2}}}{{{b^2}}}
On cross-multiplying we get,
⇒b4=a2k2\Rightarrow {b^4} = {a^2}{k^2}
Rooting both sides, we get
⇒b2=ak\Rightarrow {b^2} = ak
⇒k=b2a\Rightarrow k = \dfrac{{{b^2}}}{a}
Length of the minor axis E2{E_2}
⇒b2a=p(2b2a)\Rightarrow \dfrac{{{b^2}}}{a} = p(\dfrac{{2{b^2}}}{a})
Multiplying both the sides of the equation with ab2\dfrac{a}{{{b^2}}} , we get
⇒1=p(2)\Rightarrow 1 = p(2)
On dividing both sides by 2 we get,
p=12p = \dfrac{1}{2}

Therefore, the value of p is 12\dfrac{1}{2} .

Note: In the above solution we first make the equation of an ellipse in a standard form. An ellipse, eccentricities are the set of points in a plane in which the sum of distances from two fixed points is constant. And the length of minor-axis was also given through which we get p as 12\dfrac{1}{2} .