
If PSP’ and QSQ’ are perpendicular focal chords of a conic, then prove that
$\dfrac{1}{PS\cdot SP'}+\dfrac{1}{SQ\cdot SQ'}$ is a constant.
Answer
601.8k+ views
Hint: Use the polar form of the equation of a conic. Use the fact that if a line makes an angle y with the x-axis, then the angle made by the antiparallel line is $\pi +y$.
If the angle made by a line with the x-axis is y, then the angle made by the line perpendicular to the line is $\dfrac{\pi }{2}\pm y$.
Complete step-by-step solution -
Let the equation of the conic be $\dfrac{l}{r}=1+e\cos \theta $, where l is the length of the semi- latus rectum.
Let the angle made by the line PS with the x-axis by t.
Hence we have $\dfrac{l}{PS}=1+e\cos t$
Hence, $PS=\dfrac{l}{1+e\cos t}$
Since the angle made by PS with the x-axis is t, the angle made by the line SP’ with the x-axis is $\pi +t$
Hence, we have
$\begin{align}
& \dfrac{l}{SP'}=1-e\cos t \\
& \Rightarrow SP'=\dfrac{l}{1-e\cos t} \\
\end{align}$
Hence we have $SP'SP=\dfrac{{{l}^{2}}}{1-{{e}^{2}}{{\cos }^{2}}t}$
Since QS is perpendicular to PS, we have the angle made by QS with the x-axis is $\dfrac{\pi }{2}+t$
Hence
$\begin{align}
& \dfrac{l}{SQ}=1+e\sin t \\
& \Rightarrow SQ=\dfrac{l}{1+e\sin t} \\
\end{align}$
Since the angle made by QS with the x-axis is $\dfrac{\pi }{2}+t$, the angle made by SQ’ with the x-axis $=\dfrac{3\pi }{2}+t$
Hence, we have
$\begin{align}
& \dfrac{l}{SQ'}=1-e\sin t \\
& \Rightarrow SQ'=\dfrac{l}{1-e\sin t} \\
\end{align}$
Hence, we have $SQSQ'=\dfrac{{{l}^{2}}}{1-{{e}^{2}}{{\sin }^{2}}t}$
Hence
$\dfrac{1}{PSP'S}+\dfrac{1}{SQSQ'}=\dfrac{1-{{e}^{2}}{{\sin }^{2}}t+1-{{e}^{2}}{{\cos }^{2}}t}{{{l}^{2}}}=\dfrac{2-{{e}^{2}}}{{{l}^{2}}}$ which is a constant.
Hence proved.
Note: We can derive the polar form of the equation of a conic by considering the focus of the conic as the origin of the coordinate system and using Focal distance = eccentricity times the distance from the directrix. Then express the distance from the directrix in terms of the distance from the focus to the directrix, the eccentricity, and the angle $\theta $.
If the angle made by a line with the x-axis is y, then the angle made by the line perpendicular to the line is $\dfrac{\pi }{2}\pm y$.
Complete step-by-step solution -
Let the equation of the conic be $\dfrac{l}{r}=1+e\cos \theta $, where l is the length of the semi- latus rectum.
Let the angle made by the line PS with the x-axis by t.
Hence we have $\dfrac{l}{PS}=1+e\cos t$
Hence, $PS=\dfrac{l}{1+e\cos t}$
Since the angle made by PS with the x-axis is t, the angle made by the line SP’ with the x-axis is $\pi +t$
Hence, we have
$\begin{align}
& \dfrac{l}{SP'}=1-e\cos t \\
& \Rightarrow SP'=\dfrac{l}{1-e\cos t} \\
\end{align}$
Hence we have $SP'SP=\dfrac{{{l}^{2}}}{1-{{e}^{2}}{{\cos }^{2}}t}$
Since QS is perpendicular to PS, we have the angle made by QS with the x-axis is $\dfrac{\pi }{2}+t$
Hence
$\begin{align}
& \dfrac{l}{SQ}=1+e\sin t \\
& \Rightarrow SQ=\dfrac{l}{1+e\sin t} \\
\end{align}$
Since the angle made by QS with the x-axis is $\dfrac{\pi }{2}+t$, the angle made by SQ’ with the x-axis $=\dfrac{3\pi }{2}+t$
Hence, we have
$\begin{align}
& \dfrac{l}{SQ'}=1-e\sin t \\
& \Rightarrow SQ'=\dfrac{l}{1-e\sin t} \\
\end{align}$
Hence, we have $SQSQ'=\dfrac{{{l}^{2}}}{1-{{e}^{2}}{{\sin }^{2}}t}$
Hence
$\dfrac{1}{PSP'S}+\dfrac{1}{SQSQ'}=\dfrac{1-{{e}^{2}}{{\sin }^{2}}t+1-{{e}^{2}}{{\cos }^{2}}t}{{{l}^{2}}}=\dfrac{2-{{e}^{2}}}{{{l}^{2}}}$ which is a constant.
Hence proved.
Note: We can derive the polar form of the equation of a conic by considering the focus of the conic as the origin of the coordinate system and using Focal distance = eccentricity times the distance from the directrix. Then express the distance from the directrix in terms of the distance from the focus to the directrix, the eccentricity, and the angle $\theta $.
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