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Hyperbola

Section: Mathematics

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Question : 14 of 15

Marks: +1, -0

The circle

x62 + y^2 - 8x = 0

\frac{x^2}{9} - \frac{y^2}{4}

= 1 intersect at the points A and B.

Equation of the circle with AB as its diameter is

[JEE Adv 2010 P1]

$x^2 + y^2$ -12x + 24 = 0
$x^2 + y^2$ +12x + 24 = 0
$x^2 + y^2$ + 24x -12 = 0
$x^2 + y^2$ - 24x -12 = 0

Solution:

A point on hyperbola is (3secθ, 2tanθ).

It lies on the circle, so,

9\sec^2\theta + 4\tan^2\theta - 24\sec\theta

= 0

⇒

13\sec^2\theta - 24\sec\theta - 4

= 0 ⇒ sec θ = 2 ,

-\frac{2}{13}

Therefore,

sec θ = 2 ⇒ tan θ =

\sqrt{3}

The point of intersection are

A(6,2\sqrt{3})

and B

(6,-2\sqrt{3})

Hence, The circle with AB as diameter is

(x-6)^2 + y^2

=

(2\sqrt{3})^2

⇒

x^2 + y^2

-12x + 24 = 0

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