For the hyperbola $x^2/100 - y^2/25 = 1$, prove that $SA \cdot S'A = 25$, where S and S' are the foci and A is the vertex.

For the hyperbola x²/100 - y²/25 = 1, prove that S – Class 11 Mathematics

Miscellaneous Exercise 7(II) · Q137

x^2/100 - y^2/25 = 1

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$a^2=100,\,b^2=25 \Rightarrow a=10$ . $e^2=1+\dfrac{25}{100}=\dfrac54$ .

Let $A=(a,0)$ be the vertex nearer $S=(ae,0)$ ; $S'=(-ae,0)$ .

$SA=|a-ae|=a(e-1)$ (since $e>1$ ), and $S'A=|a-(-ae)|=a(1+e)$ .

SA\cdot S'A=a(e-1)\cdot a(e+1)=a^2(e^2-1).

But $b^2=a^2(e^2-1)$ by definition of the hyperbola, so $SA\cdot S'A=b^2$ . …