If the origin is shifted to the point (-1, 2) the new equation of the locus is $X^{2} + 5 X Y + 3 Y^{2} = 0$ , find the original equation of the locus, axes remaining parallel.

Open in App

Solution

Origin being shifted to $(- 1, 2)$ and obtained eequation is
$x^{2} + 5 x y + 3 y^{2} = 0$
For getting new from the early one we replace $x \to x + 1 = x => x = x - 1$
and $y \to y - 2 = y => y = y + 2$
$=> (x - 1)^{2} + 5 (x - 1) (y + 2) + {3 (y + 2)}^{2} = 0 => x^{2} + 8 x + 5 x y + 7 y + {3 y}^{2} + 3 = 0$

Suggest Corrections

Similar questions

(2 x - 1)^{2} + {(y + \frac{3}{2})}^{2} = 4.

(\frac{1}{2}, - \frac{3}{2})

(2, - 1)

2 x^{2} + 3 x y - 9 y^{2} - 5 x - 24 y - 7 = 0

x y - x - y + 1 = 0

x^{2} - y^{2} - 2 x + 2 y = 0

x^{2} + y^{2} - 4 x + 6 y + 3 = 0

X Y = 1

x y - 3 x + 2 y - 7 = 0

A (h, k)

h, k

x^{2} - 4 x - 6 y + 10 = 0

X^{2} + A Y = 0

Join BYJU'S Learning Program