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Question: If PSP’ and QSQ’ are perpendicular focal chords of a conic, then prove that \(\dfrac{1}{PS\cdot S...

If PSP’ and QSQ’ are perpendicular focal chords of a conic, then prove that
1PS⋅SP′+1SQ⋅SQ′\dfrac{1}{PS\cdot SP'}+\dfrac{1}{SQ\cdot SQ'} is a constant.

Explanation

Solution

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 π+y\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 π2±y\dfrac{\pi }{2}\pm y.

Complete step-by-step solution -
Let the equation of the conic be lr=1+ecos⁡θ\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 lPS=1+ecos⁡t\dfrac{l}{PS}=1+e\cos t
Hence, PS=l1+ecos⁡tPS=\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 π+t\pi +t
Hence, we have
lSP′=1−ecos⁡t ⇒SP′=l1−ecos⁡t \begin{aligned} & \dfrac{l}{SP'}=1-e\cos t \\\ & \Rightarrow SP'=\dfrac{l}{1-e\cos t} \\\ \end{aligned}
Hence we have SP′SP=l21−e2cos⁡2tSP'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 π2+t\dfrac{\pi }{2}+t
Hence
lSQ=1+esin⁡t ⇒SQ=l1+esin⁡t \begin{aligned} & \dfrac{l}{SQ}=1+e\sin t \\\ & \Rightarrow SQ=\dfrac{l}{1+e\sin t} \\\ \end{aligned}
Since the angle made by QS with the x-axis is π2+t\dfrac{\pi }{2}+t, the angle made by SQ’ with the x-axis =3π2+t=\dfrac{3\pi }{2}+t
Hence, we have
lSQ′=1−esin⁡t ⇒SQ′=l1−esin⁡t \begin{aligned} & \dfrac{l}{SQ'}=1-e\sin t \\\ & \Rightarrow SQ'=\dfrac{l}{1-e\sin t} \\\ \end{aligned}
Hence, we have SQSQ′=l21−e2sin⁡2tSQSQ'=\dfrac{{{l}^{2}}}{1-{{e}^{2}}{{\sin }^{2}}t}
Hence
1PSP′S+1SQSQ′=1−e2sin⁡2t+1−e2cos⁡2tl2=2−e2l2\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 θ\theta .