Find the new position of origin so that equation x2+4 x+8 y-2=0 will not contain a term in x and the costant term.

Question

Find the new position of origin so that equation x2+4 x+8 y-2=0 will not contain a term in x and the costant term. (A) ((3/4), 4) (B) ((3/4),-2) (C) (-2, (3/4)) (D) (-2,-(3/4))

Tardigrade Admin · Accepted Answer

x=x+h y=y+k (x+h)2+4(x+h)+8(y+k)-2=0 x2+(2 h+4) x+8 y+(h2+4 h+8 k-2)=0 ∴ 2 h+4=0 h2+4 h+8 k-2=0 h=-2 4-8+8 k-2=0 k=(6/8)=(3/4) (-2, (3/4))

Q. Find the new position of origin so that equation $x^{2} + 4 x + 8 y - 2 = 0$ will not contain a term in $x$ and the costant term.

$(\frac{3}{4}, 4)$

$(\frac{3}{4}, - 2)$

$(- 2, \frac{3}{4})$

$(- 2, - \frac{3}{4})$

$x = x + h & y = y + k$
$(x + h)^{2} + 4 (x + h) + 8 (y + k) - 2 = 0$
$x^{2} + (2 h + 4) x + 8 y + (h^{2} + 4 h + 8 k - 2) = 0$
$∴ 2 h + 4 = 0& h^{2} + 4 h + 8 k - 2 = 0$
$h = - 24 - 8 + 8 k - 2 = 0$
$k = \frac{6}{8} = \frac{3}{4}$
$(- 2, \frac{3}{4})$

Q. Find the new position of origin so that equation x2+4x+8y−2=0 will not contain a term in x and the costant term.

(43​,4)

(43​,−2)

(−2,43​)

(−2,−43​)