If the tangent at the point ( x _ 1 , y _ 1 ) on the curve y = x ^...

Given $y = {x}^{3} + 3{x}^{2} + 5$

Now $\dfrac{dy}{dx} = 3{x}^{2} + 6x$

So at point \left({x}_{1}, {y}_{1}\right)$</p><p>${\left(\dfrac{dy}{dx}\right)}_{at\left({x}_{1}, {y}_{1}\right)} = 3{x}_{1}^{2} + 6{x}_{1} . . . \left(i\right)$</p><p>Also slope of line in two point form will be$\dfrac{{y}_{1} - 0}{{x}_{1} - 0} . . . \left(ii\right)$</p><p>Also \left({x}_{1}, {y}_{1}\right)$ lie on curve

We get ${y}_{1} = {x}_{1}^{3} + 3{x}_{1}^{2} + 5 . . . \left(iii\right)$

From equation \left(i\right)$and \left(ii\right)$ we get $3{x}_{1}^{3} + 6{x}_{1}^{2} = {y}_{1}$ (as slope is equal)

Now ${y}_{1} = 3{x}_{1}^{3} + 6{x}_{1}^{2} = {x}_{1}^{3} + 3{x}_{1}^{2} + 5$

$⇒ 2{y}_{1} = 3{x}_{1}^{2} + \dfrac{15}{2}$

i.e. $2y = 3x2 + \dfrac{15}{2}$ does not intersect the curve $\dfrac{x}{3} - {y}^{2} = 2$