If $C$ is centre of a hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, $S, S'$ its foci and $P$ a point on it. Then $SP. S'P = CP^2 - a^2 + b^2$.
• A
True
• B
False

#### Solution

Given Hyperbola: $\cfrac{x^{2}}{a^{2}}-\cfrac{y^{2}}{b^{2}}=1$----------1
S & S' are foci$(ae,0)$ and $(-ae,0)$
C: Centre$(0,0)$
P: $(x_{1}, y_{1})$
a point on Hyperbola.
$\cfrac{x_{1}^{2}}{a^{2}}-\cfrac{y_{1}^{2}}{b^{2}}=1$------------2
Now calculating $S^{'}P,SP,CP$
$\overline{S^{'}P}=\sqrt{(x_{1}+ae)^{2}+(y_{1}-0)^{2}}=\sqrt{(x_{1}+ae)^{2}+(y_{1})^{2}}$--------------3
$\overline{SP}=\sqrt{(x_{1}-ae)^{2}+(y_{1}-0)^{2}}=\sqrt{(x_{1}-ae)^{2}+(y_{1})^{2}}$----------------4
$CP^{2}=\sqrt{(x_{1}-0)^{2}+(y_{1}-0)^{2}}=\sqrt{(x_{1})^{2}+(y_{1})^{2}}$-----------5
Now, $SP\cdot S^{'}P=\sqrt {[(x_{1}+ae)^{2}+y_{1}^{2}][(x_{1}-ae)^{2}+(y_{1})^{2}]}$
LHS$=\sqrt{y_{1}^{2}(x_{1}+ae)^{2}+y_{1}^{2}(x_{1}-ae)^{2}+y_{1}^{4}+(x_{1}+ae)^{2}\cdot (x_{1}-ae)^{2}}$
$=\sqrt{y_{1}^{2}(x_{1}+ae)^{2}+y_{1}^{2}(x_{1}-ae)^{2}+y_{1}^{4}+[x_{1}^{2}-a^{2}e^{2}]^{2}}$
$=\sqrt{y_{1}^{2}(2x_{1}^{2}+2a^{2}e^{2})+y_{1}^{4}+[x_{1}^{4}+a^{4}e^{4}-2x_{1}^{2}a^{2}e^{2}]}$
$=\sqrt{2x_{1}^{2}y_{1}^{2}+2a^{2}e^{2}y_{1}^{2}+y_{1}^{4}+x_{1}^{4}+a^{4}e^{4}-2x_{1}^{2}a^{2}e^{2}}$
$=\sqrt{x_{1}^{4}+2a^{2}e^{2}(y_{1}^{2}-x_{1}^{2})+a^{4}e^{4}+2x_{1}^{2}y_{1}^{2}+y_{1}^{4}}$
$=\sqrt{(x_{1}^{2}+y_{1}^{2}-a^{2}+b^{2})^{2}}$

LHS$=x_{1}^{2}+y_{1}^{2}-a^{2}+b^{2}$
RHS$=CP^{2}-a^{2}+b^{2}=x_{1}^{2}+y_{1}^{2}-a^{2}+b^{2}$-------(from Equation 5)
So, we can see LHS=RHS.
i.e. $SP\cdot S^{'}P=CP^{2}-a^{2}+b^{2}$