First slide

Cartesian plane

Question

The coordinates of the point where origin to be shifted so that the equation $y^{2} + 4 y + 8 x - 2 = 0$ will not contain the term in y and the constant term, are

Moderate

A

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

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

$(2, + \frac{3}{4})$
D

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

Solution

Let the origin be shifted to (h, k). Then, x=X+k and y=Y+k. Substituting $x = X + h, y = Y + k$ in the equation $y^{2} + 4 y + 8 x - 2 = 0$ , we get
$(Y + k)^{2} + 4 (Y + k) + 8 (X + h) - 2 = 0$
$= Y^{2} + (4 + 2 k) Y + 8 X + (k^{2} + 4 k + 8 h - 2) = 0$
For this equation to be free from the term containing Y and the constant term, we rnust have
$4 + 2 k = 0$ and $k^{2} + 4 k + 8 h - 2 = 0$
$\Rightarrow k = - 2 a$ and $h = \frac{3}{4}$ .
Hence, the origin must be shifted at the point ( 3/ 4, - 2).

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App