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A normal to the hyperbola (x^2)/(a^2)-(y...

A normal to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` meets the axes at `Ma n dN` and lines `M P` and `N P` are drawn perpendicular to the axes meeting at `Pdot` Prove that the locus of `P` is the hyperbola `a^2x^2-b^2y^2=(a^2+b^2)dot`

Text Solution

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The equation of normal at the point `Q(a sec phi, b tan phi)` to the hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
is `ax cos phi+by cot phi=a^(2)+b^(2)" (1)"`
The normal (1) meets the x-axis at
`M((a^(2)+b^(2))/(a) sec phi, 0)`
and the y-axis at
`N(0,(a^(2)+b^(2))/(b)tan phi)`
Let point P be (h,k). From the diagram,

`h=(a^(2)+b^(2))/(a)sec phi`
and `k=(a^(2)+b^(2))/(b) tan phi`
Eliminating `phi` by using the relation `sec^(2)phi-tan^(2)phi=1`, we have
`((ah)/(a^(2)+b^(2)))^(2)-((bk)/(a^(2)+b^(2)))^(2)=1`
Therefore, `a^(2)x^(2)-b^(2)y^(2)=(a^(2)+b^(2))^(2)` is the required locus of P.
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