If $S$ and $S^{'}$ be the foci, $C$ the center and $P$ be any point on a rectangular hyperbola, then $S P . S^{'} P$ is equal to :

C P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(C P)}^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{(C P)}^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(C P)}^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${(C P)}^{2}$
Rectangular hyperbola is $x^{2} - y^{2} = a^{2}$ ...(1)
$e = \sqrt{2}, C (0, 0), S (\sqrt{2} a, 0), S^{'} (- \sqrt{2} a, 0)$
Let $P (α, β)$ be any point
$P$ lies on (1)
$∴ α^{2} - β^{2} = a^{2}$ ...(2)
Now ${S P}^{2} = {(\sqrt{2} a - α)}^{2} + {(0 - β)}^{2} = 2 a^{2} + α^{2} + β^{2} - 2 \sqrt{2} a α$
${S^{'} P}^{2} = {(- \sqrt{2} a - α)}^{2} + {(0 - β)}^{2} = 2 a^{2} + α^{2} + β^{2} + 2 \sqrt{2} a α$
Now ${S P}^{2} . {S^{'} P}^{2} = {(2 a^{2} + α^{2} + β^{2})}^{2} - 8 a^{2} α^{2}$
$= 4 a^{4} + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} - 8 a^{2} α^{2} = 4 a^{2} (a^{2} - 2 α^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = 4 a^{2} (α^{2} - β^{2} - 2 α^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = - 4 a^{2} (α^{2} + β^{2}) + 4 a^{2} (α^{2} + β^{2}) + {(α^{2} + β^{2})}^{2} = {(α^{2} + β^{2})}^{2} = {((C P)^{2})}^{2} = {C P}^{4} \Rightarrow S P . S^{'} P = (C P)^{2}$

Suggest Corrections

Similar questions

C

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

S, S^{'}

P

S P . S^{'} P = C P^{2} - a^{2} + b^{2}

x^{2} - y^{2} = a^{2}

S P . S^{'} P =

P

x^{2} - y^{2} = a^{2}

C

S

S^{'}

S P \cdot S^{'} P

A rectangular hyperbola whose center is $C$ is cut by any circle of radius $r$ in four points $P, Q, R$ and $S$ . Then if,

$C P^{2} + C Q^{2} + C R^{2} + C S^{2} = k r^{2}$ , find the value of $k$ .

Join BYJU'S Learning Program