(2t2−15t+10)dt ⇒‌‌y′(x)=2x2−15x+10 Since equation of normal is parallel to x+3y=−5 ∴ Slope of normal to y(x)= Slope of lime ⇒‌‌‌
−1
[y′(x)]a,b
=‌
−1
3
or [y′(x)]a,b=3 2a2−15a+10=3 ⇒‌‌2a2−15a+7=0 ⇒‌‌(2a−1)(a−7)=0 a=‌
1
2
or 7 As, a>1, so, a=7 Now, (7,b) lies on y(x), ∴‌‌b=