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 413 along the line –3x + 2y = 1. Let H be the hyperbola x2α2−y2β2=1, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C. Then 2α2 + 3β2 is equal to ________

x – y + 1 = 0p = rα−0+12=r⇒(α+1)2=2r2…. now −3α+0−19+42+2132=r2⇒(3α+1)2+4=13r2……(2) (1) \& (2)⇒(3α+1)2+4=13(α+1)22 ⇒18α2+12α+2+8=13α2+26α+13⇒5α2−14α−3=0⇒5α2−15α+α−3=0⇒5α2−15α+α−3=0⇒α=−15,3∴r=22How αe=3 and 2α=42α2e2=9⇒α=22⇒α2=8α21+β2α2=9α2+β2=9∴β2=1∴2α2+3β2=2(8)+3(1)=19

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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 ________

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answer is 19.

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Detailed Solution

x – y + 1 = 0
p = r
$|\frac{α - 0 + 1}{\sqrt{2}}| = r \Rightarrow (α + 1)^{2} = 2 r^{2} \dots . \begin{array}{l} now {(\frac{- 3 α + 0 - 1}{\sqrt{9 + 4}})}^{2} + {(\frac{2}{\sqrt{13}})}^{2} = r^{2} \\ \Rightarrow (3 α + 1)^{2} + 4 = 13 r^{2} \dots \dots (2) \\ (1) \& (2) \Rightarrow (3 α + 1)^{2} + 4 = 13 \frac{(α + 1)^{2}}{2} \\ \Rightarrow 18 α^{2} + 12 α + 2 + 8 = 13 α^{2} + 26 α + 13 \\ \Rightarrow 5 α^{2} - 14 α - 3 = 0 \\ \Rightarrow 5 α^{2} - 15 α + α - 3 = 0 \\ \Rightarrow 5 α^{2} - 15 α + α - 3 = 0 \\ \Rightarrow α = \frac{- 1}{5}, 3 \\ ∴ r = 2 \sqrt{2} \end{array}$
$\begin{array}{l} How αe = 3 and 2 α = 4 \sqrt{2} \\ \begin{array}{l} α^{2} e^{2} = 9 \Rightarrow α = 2 \sqrt{2} \Rightarrow α^{2} = 8 \\ α^{2} (1 + \frac{β^{2}}{α^{2}}) = 9 \\ α^{2} + β^{2} = 9 \\ ∴ β^{2} = 1 \\ ∴ 2 α^{2} + 3 β^{2} = 2 (8) + 3 (1) = 19 \end{array} \end{array}$