Conjugate and Transverse Axis of Hyperbola – Length and Formula

Here you will learn formula to find the length of conjugate axis and transverse axis of hyperbola with examples.

Let’s begin –

Conjugate and Transverse Axis of Hyperbola

(i) For the hyperbola \(x^2\over a^2\) – \(y^2\over b^2\) = 1

The length of the transverse axis = 2a

And the length of the conjugate axis = 2b

(ii) For the conjugate hyperbola -\(x^2\over a^2\) + \(y^2\over b^2\) = 1

The length of the transverse axis = 2b

And the length of the conjugate axis = 2a

Also Read : Equation of the Hyperbola | Graph of a Hyperbola

Example : For the given hyperbola, find the length of conjugate axis and transverse axis.

(i)  \(16x^2 – 9y^2\) = 144

(ii)  \(9x^2 – 16y^2 – 18x + 32y – 151\) = 0

Solution :

(i)  We have,

\(16x^2 – 9y^2\) = 144 \(\implies\) \(x^2\over 9\) – \(y^2\over 16\) = 1,

This is of the form \(x^2\over a^2\) – \(y^2\over b^2\) = 1

where \(a^2\) = 9 and \(b^2\) = 16 i.e. a = 3 and b = 4

Length of the transverse axis = 2a = 6

And the length of the conjugate axis = 2b = 8

(ii) We have,

\(9x^2 – 16y^2 – 18x + 32y – 151\) = 0

\(\implies\) \(9(x^2 – 2x)\) – \(16(y^2 – 2y)\) = 151

\(\implies\)  \(9(x^2 – 2x + 1)\) – \(16(y^2 – 2y + 1)\) = 144

\(\implies\)  \(9(x – 1)^2\) – \(16(y – 1)^2\) = 144

\(\implies\)  \((x – 1)^2\over 16\) – \((y – 1)^2\over 9\) = 1

Here, a = 4 and b = 3

Length of the transverse axis = 2a = 8

And the length of the conjugate axis = 2b = 6

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