Application of Derivatives Part 2

Let the equation of normal be
(x−x1)
y−y1=(‌

−1

m

)(x−x1) . . . (i)
where, m=‌

dy1

dx1

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