Given that, it passes through point (a,b), then from Eq. (i), we get b−y1=(
−1
m
)(a−x1) =
−dx1
dy1
(a−x1) (b−y1)dy1=−(a−x1)dx1 On integrating, we get by1−
y12
2
=−ax1+
x12
2
+c . . . (ii) Eq. (ii), passes through (3,−3). and (4,−2√2), then from Eq. (ii) −3b−
9
2
=
9
2
−3a+c ⇒3a−3b−c=9. . . (iii) and −2√2b−4=8−4a+c ⇒4a−2√2b−c=12 . . . (iv) Also, given that a−2√2b=3 . . . (v) Use Eqs. (iii), (iv), (v) to solve for a,b,c, Subtract Eq. (iv) from Eq. (iii), −a+(2√2−3)b=−3. . . (vi) Now, add Eqs. (v) and (vi) b=0 Put b=0 in Eq. (vi) a=3 ∴a=3,b=0 ∴a=3,b=0 Put a=3,b=0 in a2+b2+ab a2+b2+ab=(3)2+0+0=9
