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प्रश्न
If P is any point on the hyperbola whose axis are equal, prove that SP. S'P = CP2.
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उत्तर
Equation of the hyperbola: \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
If the axes of the hyperbola are equal, then \[a = b\].
Then, equation of the hyperbola becomes \[x^2 - y^2 = a^2\].
\[\therefore b^2 = a^2 \left( e^2 - 1 \right)\]
\[ \Rightarrow a^2 = a^2 \left( e^2 - 1 \right)\]
\[ \Rightarrow 1 = \left( e^2 - 1 \right)\]
\[ \Rightarrow e^2 = 2\]
\[ \Rightarrow e = \sqrt{2}\]
Thus, the centre \[C\left( 0, 0 \right)\] and the focus are given by \[S\left( \sqrt{2}a, 0 \right)\] and \[S'\left( - \sqrt{2}a, 0 \right)\], respectively.
Let
\[P\left( \alpha, \beta \right)\] be any point on the parabola.
So, it will satisfy the equation.
\[\alpha^2 - \beta^2 = a^2\]
\[\therefore S P^2 = \left( \sqrt{2}a - \alpha \right)^2 + \beta^2 \]
\[ = 2 a^2 + \alpha^2 - 2\sqrt{2}a\alpha + \beta^2\]
\[S' P^2 = \left( - \sqrt{2}a - \alpha \right)^2 + \beta^2 \]
\[ = 2 a^2 + \alpha^2 + 2\sqrt{2}a\alpha + \beta^2\]
Now, \[S P^2 . S' P^2 = \left( 2 a^2 + \alpha^2 - 2\sqrt{2}a\alpha + \beta^2 \right)\left( 2 a^2 + \alpha^2 + 2\sqrt{2}a\alpha + \beta^2 \right)\]
\[ = 4 a^4 + 4 a^2 \left( \alpha^2 + \beta^2 \right) + \left( \alpha^2 + \beta^2 \right)^2 - 8 a^2 \alpha^2 \]
\[ = 4 a^2 \left( a^2 - 2 \alpha^2 \right) + 4 a^2 \left( \alpha^2 + \beta^2 \right) + \left( \alpha^2 + \beta^2 \right)^2 \]
\[ = 4 a^2 \left( \alpha^2 - \beta^2 - 2 \alpha^2 \right) + 4 a^2 \left( \alpha^2 + \beta^2 \right) + \left( \alpha^2 + \beta^2 \right)^2 \]
\[ = - 4 a^2 \left( \alpha^2 + \beta^2 \right) + 4 a^2 \left( \alpha^2 + \beta^2 \right) + \left( \alpha^2 + \beta^2 \right)^2 \]
\[ = \left( \alpha^2 + \beta^2 \right)^2 \]
\[ = C P^4 \]
