If the equation $x^{2} - 4 x - 6 y + 10 = 0$ is transformed to $X^{2} + A Y = 0$ axes remaining parallel. Find the co-ordinates of the point where the origin is shifted and value of A.

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Solution

The old equation is,

$x^{2} - 4 x - 6 y + 10 = 0$ ..........(1)

Let (x,y) be the old coordinates and (X,Y) be the new coordinates after shifting the origin to (h,k)

we have, $x = X + h$ and $y = Y + k$

$(X + h)^{2} - 4 (X + h) - 6 (Y + k) + 10 = 0$

$X^{2} + 2 X h + h^{2} - 4 X - 4 h - 6 k + 10 = 0$

$X^{2} + (2 h - 4) X - 6 Y + h^{2} - 4 h - 6 k + 10 = 0$ .........(2)

The new equation is $X^{2} + A Y = 0$

Thus terms of x and constants are absent

$2 h - 4 = 0$

$\Rightarrow h = 2$

Similarly,

$h^{2} - 4 h - 6 k + 10 = 0$

$- 6 k = - 6$

$\Rightarrow k = 1$

Thus the origin is shifted to (2,1)

Thus the equation (2) reduces to
$X^{2} - 6 Y = 0$

Comparing with given new equation of locus, A $= - 6$

Origin is shifted to $(2, 1)$ and A $= - 6$

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