The number of points on the curve y = 54 x ^ 5 - 135 x ^ 4 - 70 x ^...

Given curve is $y = 54{x}^{5} - 135{x}^{4} - 70{x}^{3} + 180{x}^{2} + 210x$

Normal lines on this curve, are parallel to $x + 90y + 2 = 0$, hence slope of normal is $m = - \dfrac{1}{90}$, so slope of tangent to the given curve is $- \dfrac{1}{m} = 90$.

Slope of tangent on the given curve is

$\dfrac{dy}{dx} = 270{x}^{4} - 540{x}^{3} - 210{x}^{2} + 360x + 210$

Here, $\dfrac{dy}{dx} = 90$

$⇒ 270{x}^{4} - 540{x}^{3} - 210{x}^{2} + 360x + 210 = 90$

$⇒ 270{x}^{4} - 540{x}^{3} - 210{x}^{2} + 360x + 120 = 0$

$⇒ 27{x}^{4} - 54{x}^{3} - 21{x}^{2} + 36x + 12 = 0$

$⇒ 9{x}^{2} - 18{x}^{3} - 7{x}^{2} + 12x + 4 = 0$

$⇒ ( x - 1 ) ( 9{x}^{3} - 9{x}^{2} - 16x - 4 ) = 0$

$⇒ ( x - 1 ) ( x - 2 ) ( 9{x}^{2} + 9x + 2 ) = 0$

$⇒ ( x - 1 ) ( x - 2 ) ( 3x + 2 ) ( 3x + 1 ) = 0$

$⇒ x = 1, 2, - \dfrac{2}{3}, - \dfrac{1}{3}$

Hence, there are four points.