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प्रश्न
Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola `"x"^2/16 - "y"^2/36 = "k"`.
उत्तर
The equation of the hyperbola is
`"x"^2/16 - "y"^2/36 = "k"` i.e. `"x"^2/"16 k" - "y"^2/"36k" = 1`
Comparing this equation with `"x"^2/"a"^2 - "y"^2/"b"^2 = 1`, we get
a2 = 16k, b2 = 36k
∴ a = `4sqrt"k", "b" = 6sqrt"k"`
∴ l(transverse axis) = 2a = `8sqrt"k"`
and l(conjugate axis) = 2b = `12sqrt"k"`
Let 2A and 2B be the lengths of the transverse and conjugate axes of the required hyperbola.
Then according to the given condition
2A = a = `4sqrt"k" and 2"B" = "b" = 6sqrt"k"`
∴ A = `2sqrt"k"` and B = `3sqrt"k"`
∴ equation of the required hyperbola is
`"x"^2/"A"^2 - "y"^2/"B"^2 = 1`
i.e. `"x"^2/"4k" - "y"^2/"9k" = 1`
∴ 9x2 - 4y2 = 36k, where k is an arbitrary constant.
Differentiating w.r.t. x, we get
`9 xx "2x" - 4 xx "2y" "dy"/"dx" = 0`
∴ `"9x" - "4y" "dy"/"dx" = 0`
This is the required D.E.
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