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. $$SP\cdot S^{'}P=CP^{2}-a^{2}+b^{2}$$