When the origin shifted to a suitable point p, the equation 2x^2 + y^2 - 4x +4y=0 transformed as 2x^2 + y^2 - 8x + 8y + 18 = 0 then p=
Answers
Answer:
The point P can be (-1, -6) or (-1, 2)
Step-by-step explanation:
Given
Original equation
2x^2+y^2-4x+4y=02x
2
+y
2
−4x+4y=0 ...... (1)
The transformed equation in new coordinate system (X,Y)
2X^2+Y^2-8X+BY+18=02X
2
+Y
2
−8X+BY+18=0 ......... (2)
Let the point P be (h,k)
Then in new coordinate system
X=x-hX=x−h or x=X+hx=X+h
Y=y-kY=y−k or y=Y+ky=Y+k
Therefore, from (2)
2(x-h)^2+(y-k)^2-8(x-h)+B(y-k)+18=02(x−h)
2
+(y−k)
2
−8(x−h)+B(y−k)+18=0
\implies 2(x^2-2xh+h^2)+(y^2-2yk+k^2)-8x+8h+By-Bk+18=0⟹2(x
2
−2xh+h
2
)+(y
2
−2yk+k
2
)−8x+8h+By−Bk+18=0
\implies 2x^2+y^2-(4h+8)x+(-2k+B)y+(2h^2+k^2+8h-Bk+18)=0⟹2x
2
+y
2
−(4h+8)x+(−2k+B)y+(2h
2
+k
2
+8h−Bk+18)=0
Comparing this equation with equation (1)
4h+8=44h+8=4
\implies 4h=-4⟹4h=−4
\implies h=-1⟹h=−1
And
-2k+B=4−2k+B=4
\implies B=4+2k⟹B=4+2k
Also,
(2h^2+k^2+8h-Bk+18)=0(2h
2
+k
2
+8h−Bk+18)=0
\implies 2(-1)^2+k^2+8(-1)-(4+2k)k+18=0⟹2(−1)
2
+k
2
+8(−1)−(4+2k)k+18=0
\implies 2+k^2-8-4k-2k^2+18=0⟹2+k
2
−8−4k−2k
2
+18=0
\implies -k^2-4k+12=0⟹−k
2
−4k+12=0
\implies k^2+4k-12=0⟹k
2
+4k−12=0
\implies k^2+6k-2k-12=0⟹k
2
+6k−2k−12=0
\implies k(k+6)-2(k+6)=0⟹k(k+6)−2(k+6)=0
\implies (k+6)(k-2)=0⟹(k+6)(k−2)=0
\implies k=-6, 2⟹k=−6,2
Therefore, the twp possible values of P can be
(-1,-6)(−1,−6) and (-1,2)(−1,2)
Hope this answer is helpful.
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