Question

Solve the Question !

Shift the origin to a suitable point so that the equation y² + 4y +8X - 2 =0 will not contain a term y and the constant term .
​

Answer 1

Answer:

k=-2 and h =3/h

Solution :- Let origin ne shifted to (h,k) . then

x = X +h and y = Y+k

substituting x =X +h and y = Y+k in the equation y² + 4y +8x -2 =0 we get ,

(Y+k)² + 4(Y+K) + 8(X+h) -2 =0

or, Y² + (4 +2k)Y +8X +(k²+4k+8h -2) =0

for this equation to be free from the term contains Y and the constant term ,we must have

4+2k = 0 ,k = -2 and k²+4k +8h -2 =0 =>k = 3/4

hence, the origin is shifted at the point (3/4,-2)

________________________________

Hope it helps !;

@Raj ,❤

Answer 2

Step-by-step explanation:

Let the suitable point be P(h,k).

Then,

x = x + h and y = y + k.

Substitute in the given equation, we get

=> (y + k)² + 4(y + k) + 8(x + h) - 2 = 0

=> y² + k² + 2ky + 4y + 4k + 8x + 8h - 2 = 0

=> y² + (4 + 2k)y + 8x + (k² + 4k + 8h - 2) = 0

Given, It will not contain a term y and the constant term.

=> (4 + 2k) = 0 and k² + 4k + 8h - 2

=> k = -2 and k = 3/4

Therefore,

The suitable point is (3/4, -2)

Hope it helps!