If A and B are the points of intersection of the circle and the hyperbola and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of PAB lies on the line : [2025]

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Solution

Q.
If A and B are the points of intersection of the circle $x^{2} + y^{2} - 8 x = 0$ and the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ and a point P moves on the line 2x – 3y + 4 = 0, then the centroid of $△$ PAB lies on the line : [2025]

1 x + 9y = 36

2 9x – 9y = 32

3 4x – 9y = 12

4 6x – 9y = 20

Ans.
(4)

We have, $x^{2} + y^{2} - 8 x = 0$ ... (i)

and $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1 \Rightarrow 4 x^{2} - 9 y^{2} = 36$ ... (ii)

Solving (i) and (ii), we get

$4 x^{2} - 9 (8 x - x^{2}) = 36$

$\Rightarrow 13 x^{2} - 72 x - 36 = 0$

$\Rightarrow (13 x + 6) (x - 6) = 0$

$\Rightarrow x = \frac{- 6}{13}, 6$

$∴ x = 6$ $[∵ x = \frac{- 6}{13} not possible]$

From (i), $y^{2} = 8 (6) - {(6)}^{2} = 48 - 36 = 12 \Rightarrow y = \pm \sqrt{12}$

$∴$ $(6, \sqrt{12})$ and $(6, - \sqrt{12})$ are the points of intersection of circle and hyperbola.

Let $P (α, β)$ be the point moves on the line 2x – 3y + 4 = 0 such that

$2 α - 3 β + 4 = 0$ ... (iii)

Centroid of $△$ PAB is given by $(\frac{α + 6 + 6}{3}, \frac{β}{3}) = (h, k)$

$∴ α = 3 h - 12 and β = 3 k$

From (iii), 2(3h –12) – 3(3k) + 4 = 0

$\Rightarrow 6 h - 24 - 9 k + 4 = 0 \Rightarrow 6 h - 9 k = 20$

i.e., 6x – 9y = 20.

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