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A normal to the hyperbola, 4x^29y^2 = 36...

A normal to the hyperbola, `4x^2_9y^2 = 36` meets the co-ordinate axes `x` and `y` at `A` and `B.` respectively. If the parallelogram `OABP (O` being the origin) is formed, then the locus of `P` is :-

Text Solution

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The correct Answer is:
`(9)/(4x^(2))-(1)/(y^(2))=1`
We have hyperbola
`(x^(2))/(9)-(y^(2))/(4)=1`
Equation of tangent to hyperbola at point `P(3 sec theta, 2 tan theta)` is given by
`(x)/(3)sec theta-(y)/(2)tan theta=1`
It meets the axes at `A(3 cos theta, 0) and B(0, -2 cot theta)`.
Let the midpoint of AB be (h, k). Then
`2h=2 cos theta and 2k=-2 cot theta`
`therefore" "sec theta=(3)/(2h) and tan theta=-(1)/(k)`
Squaring and subtracting, we get
`(9)/(4h^(2))-(1)/(k^(2))=1`
So, the locus of AB is `(9)/(4x^(2))-(1)/(y^(2))=1`.
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