find common tangent between curves $y=x^3$ and $112x^2 +y^2 =112$

Question

Find common tangent between curves $y=x^3$ and $112x^2 +y^2 =112$

For common tangent slope of both the curve must be equal and then we have to find the point of intersection of the curves.

Problem is that I can't able to find the point of intersection of the curves.

Answer 1

$y=mx+n$ is a tangent to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ iff $n^2=a^2m^2+b^2$.

Let $(t,t^3)$ is touching point of the tangent to the graph of $y=x^3$.

Thus, $y-t^3=3t^2(x-t)$ or $y=3t^2x-2t^3$ is an equation of the common tangent.

Thus, $$(-2t^3)^2=1\cdot(3t^2)^2+112.$$ Can you end it now?

Answer 2

To find a common tangent line between the two curves means to find a line that is tangent to both curves. By means of (implicit) differentiation you can find the slope of each curve at each point, and from there you can determine the tangent line at each point.

This gives you two collections of tangent lines; the two sets of tangent lines of the two curves. Now it remains to find lines that are in both collections; these are the common tangents between the curves.

Answer 3

The tangent of $y=x^3$ at $(x_0,y_0)$: $$ y= 3 x_0 ^2 x - 2 x_0 ^3 \tag1 $$ the tangent of $ 112 x^2 + y^2 = 112$ at $(x_1, y_1)$: $$ 112 x_1 x + y_1 y =112 \tag2 $$ and an equation $$ 112 x_1 ^2 + y_1 ^2 =112 \tag3 $$ Combine formula $(1),(2),(3)$ to solve $x_0,x_1,y_1$. the slope is the same, so $$ \frac{-112x_1}{y_1} = 3 x_0 ^2 \tag4 $$ combine $2$ and $4$: $$ y - 3 x_0 ^2 x = \frac{112}{y_1} \tag5 $$ so we have $$ \begin{align} x_1 &= \frac{3}{2x_0}\\ y_1 &= \frac{-56}{x_0 ^3} \end{align} $$ substitute $3$ and obtain $$ (x_0 ^2 -4)[4 x_0 ^4 + 7x_0 ^2 +28 ] =0 $$ the solution in real number is $x_0 = \pm 2$, so the tangent lines are $$ y = 12x \pm 16 $$