Statement-I: The point to which the origin has to be shifted to eliminate x and y terms in the equation $a {(x + α)}^{2} + b {(y + β)}^{2} = c i s (- α, - β)$
Statement-II : The point to which the origin has to be shifted to eliminate x and y terms in $a x^{2} + b y^{2} + 2 g x + 2 f y + c = 0 i s (\frac{- g}{a}, \frac{- f}{b})$
Statement-III: If the axes are rotated through an angle $90^{0}$ then transformed equation of $x^{2} - y^{2} = 1 i s x^{2} - y^{2} = - 1$
which of the above statements are true

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I, II are true

II, III are true

III, I are true

All are true

answer is D.

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Detailed Solution

$Let (h, k) be required point to which the origin shifted . Let (x, y) be original coordinates and (X, Y) be new coordinates . then x = X + h, y = Y + k Given equation a {(x + α)}^{2} + b {(y + β)}^{2} = c \Rightarrow a {(X + h + α)}^{2} + b {(Y + k + β)}^{2} = c \Rightarrow a (X^{2} + h^{2} + α^{2} + 2 h X + 2 hα + 2 X α) + b (Y^{2} + k^{2} + β^{2} + 2 Y k + 2 kβ + 2 Y β) = c to eliminate X and Y terms i n t h e a b o v e equating their coefficients s h o u l d b e zero . \Rightarrow a (2 h + 2 α) = 0 and b (2 k + 2 β) = 0 \Rightarrow h + α = 0 and k + β = 0 \Rightarrow h = - α and k = - β ∴ Required point = (h, k) = (- α, - β) ∴ Statement - I is true$

$Statement - II Let (h, k) be origin shifted point then x = X + h, y = Y + k Given original equation {ax}^{2} + {by}^{2} + 2 gx + 2 fy + c = 0 Now transformed equation is a {(X + h)}^{2} + b {(Y + k)}^{2} + 2 g (X + h) + 2 f (Y + k) + c = 0 to eliminate X and Y terms, its coefficients should be zero \begin{matrix} \Rightarrow 2 ah + 2 g = 0 \\ \Rightarrow 2 ah = - 2 g \\ \Rightarrow h = \frac{- g}{a} \end{matrix} \begin{matrix} and 2 bk + 2 f = 0 \\ \Rightarrow 2 b k = - 2 f \\ \Rightarrow k = \frac{- f}{b} \end{matrix} ∴ Required point (h, k) = (\frac{- g}{a}, \frac{- f}{b}) ∴ Statement - II true .$

$Statement - III Given θ = 90^{0} then x = Xcos 90^{0} - Y \sin 90^{0} \Rightarrow x = - Y and y = X s in 90^{0} + Y \cos 90^{0} y = X Now transformed equation o f x^{2} - y^{2} = 1 i s Y^{2} - X^{2} = 1 \Rightarrow X^{2} - Y^{2} = - 1 Statement - III is also true .$