Let the circle C touch the line x – y + 1 = 0, have the centre on the positive x-axis, and cut off a chord of length along the line –3x + 2y = 1. Let H be the hyperbola , whose one of the foci is the centre of C and the length of the transverse axis is t

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Solution

Q.
Let the circle C touch the line x – y + 1 = 0, have the centre on the positive x-axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line –3x + 2y = 1. Let H be the hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ , whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then $2 α^{2} + 3 β^{2}$ is equal to __________. [2025]

Ans.
(19)

Since, centre of circle lies on positive x-axis and one of the foci of hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ are same.

$∴$ Centre of circle = $(α, 0)$

Since, x – y + 1 = 0 is tangent to the circle.

$∴$ $| \frac{α + 1}{\sqrt{2}} |$ $= r$ [where 'r' is the radius of circle]

$\Rightarrow {(α + 1)}^{2} = 2 r^{2}$ ... (i)

Also, –3x + 2y = 1 is the chord of the circle

$∴$ $| \frac{- 3 α - 1}{\sqrt{9 + 4}} |^{2}$ ${+ (\frac{2}{\sqrt{13}})}^{2} = r^{2}$ $[∵ Length of chord = \frac{4}{\sqrt{13}}]$

$∴ \frac{{(3 α + 1)}^{2}}{13} + \frac{4}{13} = r^{2}$

$\Rightarrow \frac{9 α^{2} + 1 + 6 α + 4}{13} = \frac{(α^{2} + 1 + 2 α)}{2}$

$\Rightarrow 5 α^{2} - 14 α - 3 = 0$

$\Rightarrow α = 3$ [ $∵ α > 0$ ]

$∴ r = 2 \sqrt{2}$

Now, $α e = 3 and 2 α = 4 \sqrt{2}$

$∴ α^{2} e^{2} = 9 and α = 2 \sqrt{2}$

$\Rightarrow α^{2} (1 + \frac{β^{2}}{α^{2}}) = 9 \Rightarrow 8 + β^{2} = 9 \Rightarrow β^{2} = 1$

$∴ 2 α^{2} + 3 β^{2} = 2 (8) + 3 \times 1 = 16 + 3 = 19$ .

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