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If the normal at a point P to the hyperb...

If the normal at a point `P` to the hyperbola `x^2/a^2 - y^2/b^2 =1` meets the x-axis at `G`, show that the `SG = eSP.S` being the focus of the hyperbola.

Text Solution

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Equation of normal at point `P(x_(1),y_(1))` is
`(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)e^(2)`
It meets x-axis at `G(e^(2)x_(1),0)`.

Now, GL is perpendicular to the asymptote `bx=ay=0`.
So, equation of GL is
`ax+by=ae^(2)x_(1)`
Solving this line with asymptote `bx-ay=0,` we get `x=x_(1)` for point L.
Thus, abscissa of P and L are the same.
Therefore, LP is parallel to conjugate axis.
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