Question

If \[P(x,y)\] is any point on the ellipse \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]\[a>b\], and \[{{S}_{1,}}{{S}_{2}}\] are the foci, then prove that \[P{{S}_{1}}+P{{S}_{2}}=2a\].

Answer 1

Hint: As, in question it is given that a > b, so, x – axis will be the major axis. So, the foci of this ellipse are given as \[{{S}_{1}}(ae,0)\] and \[{{S}_{2}}(-ae,0)\]and ‘e’ is the eccentricity of the ellipse, which is given as, \[e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}\] . so, what we will do is we will find the distance of lines $P{{S}_{1}}$ and $P{{S}_{2}}$using general equation of an ellipse and e and then we will add the length which will leads to 2a.

Complete step-by-step answer:
The given equation of ellipse is \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\], by rewriting it we have:
\[\dfrac{{{b}^{2}}{{x}^{2}}}{{{a}^{2}}{{y}^{2}}}+\dfrac{{{a}^{2}}{{y}^{2}}}{{{a}^{2}}{{b}^{2}}}={{a}^{2}}{{b}^{2}}\]
\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
Therefore, \[{{y}^{2}}=\left( 1-\dfrac{{{x}^{2}}}{{{a}^{2}}} \right){{b}^{2}}\]
The foci of this ellipse are given as \[{{S}_{1}}(ae,0)\] and \[{{S}_{2}}(-ae,0)\],
Where ‘e’ is the eccentricity of the ellipse, which is given as,
\[e=\sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}\] …….(1)
Now, the distance between point \[P(x,y)\] and \[{{S}_{1}}(ae,0)\] is:
\[P{{S}_{1}}=\sqrt{{{\left( x-ae \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\]
By substituting, \[{{y}^{2}}=\left( 1-\dfrac{{{x}^{2}}}{{{a}^{2}}} \right){{b}^{2}}\] in the above equation we will have:
\[=\sqrt{{{x}^{2}}+{{a}^{2}}{{e}^{2}}-2aex+\left( 1-\dfrac{{{x}^{2}}}{{{a}^{2}}} \right)}{{b}^{2}}\]
\[=\sqrt{{{e}^{2}}\left( {{x}^{2}}+\dfrac{{{a}^{2}}}{{{e}^{2}}}-2\dfrac{a}{e}x \right)}\]
\[=\sqrt{{{e}^{2}}{{x}^{2}}+{{a}^{2}}-2aex}\]
\[=e\left( \dfrac{a}{e}-x \right)\]
Therefore, \[P{{S}_{1}}=a-ex\].
Now the distance between \[P(x,y)\]and \[{{S}_{2}}(-ae,0)\] is given as,
\[P{{S}_{2}}=\sqrt{{{\left( x+ae \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\]
By substituting, \[{{y}^{2}}=\left( 1-\dfrac{{{x}^{2}}}{{{a}^{2}}} \right){{b}^{2}}\] in the above equation we will have:
\[=\sqrt{{{x}^{2}}+{{a}^{2}}{{e}^{2}}+2aex+\left( 1-\dfrac{{{x}^{2}}}{{{a}^{2}}} \right){{b}^{2}}}\]
\[=\sqrt{{{e}^{2}}{{x}^{2}}+{{a}^{2}}+2aex}\]
\[=\sqrt{{{e}^{2}}({{x}^{2}}+\dfrac{{{a}^{2}}}{{{e}^{2}}}+2\dfrac{a}{e}x)}\]
\[P{{S}_{2}}=a+xe\]
\[P{{S}_{2}}=a+xe\].
Now, \[P{{S}_{1}}+P{{S}_{2}}=\left( a+xe \right)+\left( a-xe \right)\]\[=2a\].
Thus, we have proved that \[P{{S}_{1}}+P{{S}_{2}}=2a\] when \[p(x,y)\] is any point on the ellipse \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]\[a>b\],and \[{{S}_{1,}}{{S}_{2}}\] are the foci of the ellipse.

Note: The sum of focal distances from a point on the ellipse is always equal to the length of the major axis of the given ellipse. This can be remembered as a fact for any given ellipse curve. Also, while solving such questions, the concept of ellipse and foci must be remembered and calculation should be done accurately as this may lead to incorrect answers.