Question

Find the equation of the tangent at (2,3) on the curve $y^2=a{x}^3+b$ is $y=4x-5$. Find the value of a and b

Answer 1

$y^2=a{x}^3+b2y\frac{dy}{dx}=3a{x}^{2}at(2,3)6\frac{dy}{dx}=3a(2{)}^2=12a\therefore a=\frac{1}{2}\frac{dy}{dx}$
Since equation of tangent is y=4x-5
slope $=4-\frac{dy}{dx}=2a\therefore a=2$
$\therefore$ eqn: of curve $y^2=2x^3+b$
Point (2,3)is on curve
$\therefore 3^2=2(2{)}^3+b\Rightarrow 9=1.6+b$
or $b=-7$
$a=2\&b=-7$