Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that

Miscellaneous Exercise 6(II) · Q130

\dfrac{x^2}{16}-\dfrac{y^2}{36}=k

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The given hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{36}=k$ , i.e. $\dfrac{x^2}{16k}-\dfrac{y^2}{36k}=1$ , has semi-transverse axis $4\sqrt k$ and semi-conjugate axis $6\sqrt k$ , so full transverse $=8\sqrt k$ and full conjugate $=12\sqrt k$ . A hyperbola with HALF these lengths has semi-transverse $2\sqrt k$ and semi-conjugate $3\sqrt k$ : $\dfrac{x^2}{4k}-\dfrac{y^2}{9k}=1$ , i.e. $\dfrac{x^2}{4}-\dfrac{y^2}{9} …

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Q.Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{36}=k$ .

Concept understanding — Formation of a Differential Equation whose General Solution is Given

Q.Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola x216−y236=k\dfrac{x^2}{16}-\dfrac{y^2}{36}=k16x2​−36y2​=k.

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Q.Form the differential equation of the hyperbola whose length of transverse and conjugate axes are half of that of the given hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{36}=k$ .

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