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Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola xykx216-y236=k

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Question

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"`.

Sum
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Solution

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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Chapter 6: Differential Equations - Miscellaneous exercise 2 [Page 217]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 6 Differential Equations
Miscellaneous exercise 2 | Q 4.5 | Page 217

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