The equation of the common tangents to the two hyperbolas {x²}{a²} - {y²}{b²} = 1 and {y²}{a²} - {x²}{b²} = 1 ,are

Correct Answer is : y=±x±a2−b2y = ± x ± {a² - b²}y=±x±a2−b2

Correct Answer is : y=±x±a2−b2y = ± x ± {a² - b²}y=±x±a2−b2

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VITEEE 2010 Paper

Question : 105 of 120

Marks: +1, -0

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

,are

Solution:

We have the two hyperbolas as

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

...(i)

and

\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

...(ii)

Any tangent to the hyperbola eq.(i)

y = mx + c

where

c = \pm \sqrt{a^{2}m^{2} - b^{2}}

...(iii)

But this tangent touches the parabola eq.(ii) also

∴

\frac{(mx + c)^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

⇒

b^{2}(m^{2}x^{2} + c^{2} + 2mcx) - a^{2}x^{2} = a^{2}b^{2}

⇒

(b^{2}m^{2} - a^{2})x^{2} + 2mcb^{2}x + b^{2}(c^{2} - a^{2}) = 0

For the tangency, it should have equal roots

(2mcb^{2})^{2} = 4(b^{2}m^{2} - a^{2}) \cdot b^{2}(c^{2} - a^{2})

⇒

4m^{2}c^{2}b^{4} = 4b^{2}(b^{2}m^{2}c^{2} - b^{2}m^{2}a^{2} - a^{2}c^{2} + a^{4})

⇒

c^{2} = a^{2} - b^{2}m^{2}

⇒

a^{2}m^{2} - b^{2} = a^{2} - b^{2}m^{2} \qquad [\text{using Eq. (iii)}]

⇒

(a^{2} + b^{2})m^{2} = a^{2} + b^{2}

⇒

m^{2} = 1 \Rightarrow m = \pm 1

Hence, the equation of common tangent are

y = \pm x \pm \sqrt{a^{2} - b^{2}}

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