Answer
a) $y=\frac{9}{2}x-\frac{5}{2}$
b) See image below.
Work Step by Step
a)
$y^2=5x^4-x^2\\
2yy'=20x^3-2x\\
yy'=10x^3-x\\
y'=\frac{10x^3-x}{y}$
Plug in $(1,2)$ into $y′$ to find the gradient
$y'=\frac{10(1)^3-(1)}{2}\\
y'=\frac{9}{2}$
Equation of the tangent line at $(1,2)$
$y-2=\frac{9}{2}(x-1)\\
y=\frac{9}{2}x-\frac{5}{2}$
b) See image.