Question

If $C$ is centre of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $S,S^{\prime }$ its foci and $P$ a point on it. Then $SP.S^{\prime }P=C{P}^2-a^2+b^2$.

• A. True
• B. False

Answer 1

The correct option is A True

Given Hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$----------1

S & S' are foci$(ae,0)$ and $(-ae,0)$

C: Centre$(0,0)$

P: $(x_1,y_1)$

a point on Hyperbola.

$\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$------------2

Now calculating $S^{{}^{\prime }}P,SP,CP$

$\overline{S^{{}^{\prime }}P}=\sqrt{(x_1+ae{)}^2+(y_1-0{)}^2}=\sqrt{(x_1+ae{)}^2+(y_1{)}^2}$--------------3

$\overline{SP}=\sqrt{(x_1-ae{)}^2+(y_1-0{)}^2}=\sqrt{(x_1-ae{)}^2+(y_1{)}^2}$----------------4

$C{P}^2=\sqrt{(x_1-0{)}^2+(y_1-0{)}^2}=\sqrt{(x_1{)}^2+(y_1{)}^2}$-----------5

Now, $SP\cdot S^{{}^{\prime }}P=\sqrt{\left[\right(x_1+ae{)}^2+y_1^2\left]\right[(x_1-ae{)}^2+(y_1{)}^2]}$

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$=C{P}^2-a^2+b^2=x_1^2+y_1^2-a^2+b^2$-------(from Equation 5)

So, we can see LHS=RHS.

i.e. $SP\cdot S^{{}^{\prime }}P=C{P}^2-a^2+b^2$