
At some point P on the ellipse, the segment SS’ subtends a right angle, then its eccentricity is
( A ) \[e = \dfrac{{\sqrt 2 }}{2}\]
( B ) \[e < \dfrac{1}{{\sqrt 2 }}\]
( C ) \[e > \dfrac{1}{{\sqrt 2 }}\]
( D ) \[e = \dfrac{{\sqrt 3 }}{2}\]
Answer
594.6k+ views
Hint: To deal with this sort of problem we are going to utilize the below steps so that we can make our solution as straightforward as possible and that will be useful to spare our significant time.
We will assume a point \[P(a\cos \theta ,b\sin \theta )\]
Then will find the slope of SP and S’P and by using product of slope = -1
We will get eccentricity.
Given: We are given that at some point P on the ellipse, the segment SS’
subtends a right angle
Complete step-by-step answer:
When we are solving this type of question, we need to follow the steps provided in the hint part above.
\[\begin{array}{l}
P(a\cos \theta ,b\sin \theta )\\
as\,\,S = (ae,0)\\
S' = ( - ae,0)\\
slope\,\,of\,\,PS({m_1}) = \dfrac{{bsin\theta - 0}}{{a\cos \theta - ae}}\\
slope\,\,of\,\,PS'({m_2}) = \dfrac{{bsin\theta - 0}}{{a\cos \theta + ae}}
\end{array}\]
\[\begin{array}{l}
as\,\,{m_1}{m_2} = - 1\\
\dfrac{{{b^2}si{n^2}\theta }}{{{a^2}{{\cos }^2}\theta - {a^2}{e^2}}} = - 1\\
as\,\,e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \\
{b^2} = {a^2}(1 - {e^2})\\
so\\
\Rightarrow {a^2}(1 - {e^2})si{n^2}\theta + {a^2}{\cos ^2}\theta - {a^2}{e^2} = 0\\
\Rightarrow {a^2}({\cos ^2}\theta + si{n^2}\theta ) + {a^2}{e^2}(si{n^2}\theta + 1) = 0\\
\Rightarrow 1 = {e^2}(si{n^2}\theta + 1)\\
\Rightarrow {e^2} = \dfrac{1}{{si{n^2}\theta + 1}}\\
\Rightarrow e = \dfrac{1}{{\sqrt {si{n^2}\theta + 1} }}\\
as\,\,minimum\,\,value\,\,of\,\,\dfrac{1}{{\sqrt {si{n^2}\theta + 1} }}\,\,is\dfrac{1}{{\sqrt 2 }}\\
So\,\,e > \dfrac{1}{{\sqrt 2 }}
\end{array}\]
Hence, we can clearly say that option (C) is the correct answer.
Hence after following the each and every step given in the hint part we obtained our final answer.
Additional Information:
Here we can clearly see that in this solution we did not use any complicated process because we followed basic and simple things in the right order as per given in the above hint section.
Note:
In this type of question, we need to take care of the many things and some of them are mentioned here which will be really helpful to understand the concept:
We need to use the correct formula in such a way that the solution does not become too complex.
Product of slope is =-1
We will assume a point \[P(a\cos \theta ,b\sin \theta )\]
Then will find the slope of SP and S’P and by using product of slope = -1
We will get eccentricity.
Given: We are given that at some point P on the ellipse, the segment SS’
subtends a right angle
Complete step-by-step answer:
When we are solving this type of question, we need to follow the steps provided in the hint part above.
\[\begin{array}{l}
P(a\cos \theta ,b\sin \theta )\\
as\,\,S = (ae,0)\\
S' = ( - ae,0)\\
slope\,\,of\,\,PS({m_1}) = \dfrac{{bsin\theta - 0}}{{a\cos \theta - ae}}\\
slope\,\,of\,\,PS'({m_2}) = \dfrac{{bsin\theta - 0}}{{a\cos \theta + ae}}
\end{array}\]
\[\begin{array}{l}
as\,\,{m_1}{m_2} = - 1\\
\dfrac{{{b^2}si{n^2}\theta }}{{{a^2}{{\cos }^2}\theta - {a^2}{e^2}}} = - 1\\
as\,\,e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \\
{b^2} = {a^2}(1 - {e^2})\\
so\\
\Rightarrow {a^2}(1 - {e^2})si{n^2}\theta + {a^2}{\cos ^2}\theta - {a^2}{e^2} = 0\\
\Rightarrow {a^2}({\cos ^2}\theta + si{n^2}\theta ) + {a^2}{e^2}(si{n^2}\theta + 1) = 0\\
\Rightarrow 1 = {e^2}(si{n^2}\theta + 1)\\
\Rightarrow {e^2} = \dfrac{1}{{si{n^2}\theta + 1}}\\
\Rightarrow e = \dfrac{1}{{\sqrt {si{n^2}\theta + 1} }}\\
as\,\,minimum\,\,value\,\,of\,\,\dfrac{1}{{\sqrt {si{n^2}\theta + 1} }}\,\,is\dfrac{1}{{\sqrt 2 }}\\
So\,\,e > \dfrac{1}{{\sqrt 2 }}
\end{array}\]
Hence, we can clearly say that option (C) is the correct answer.
Hence after following the each and every step given in the hint part we obtained our final answer.
Additional Information:
Here we can clearly see that in this solution we did not use any complicated process because we followed basic and simple things in the right order as per given in the above hint section.
Note:
In this type of question, we need to take care of the many things and some of them are mentioned here which will be really helpful to understand the concept:
We need to use the correct formula in such a way that the solution does not become too complex.
Product of slope is =-1
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