AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

Equation of given curve is
3y2=2ax2+6b ....(i)
∵ Curve (i) passes through point P(3,−1), so
3=18a+6b
⇒‌‌6a+2b=1...(ii)
and the ‌

dy

dx

(3,−1)=−1 (given), so on differentiating the Eq. (i), we get
6y‌

dy

dx

=4ax
∴b=−1{ from Eq. (ii) on putting value of ' a′}