Question

The equations of the tangents to the curve $y=x^4$ from the point $(2,0)$ not on the curve, are given by

• A. $y=0$
• B. $y-1=5(x-1)$
• C. $y-\frac{4098}{81}=\frac{2048}{27}(x-\frac{8}{3})$
• D. $y-\frac{32}{243}=\frac{80}{81}(x-\frac{2}{3})$

Answer 1

Answer: (A), (C)

The correct options are
A $y=0$
C $y-\frac{4098}{81}=\frac{2048}{27}(x-\frac{8}{3})$

Let $(x_0,x_0^4)$ be the point of tangency.
Then the equation of the tangent will be $y-x_0^4=y\left(x_0\right)(x-x_0)$. Since this tangent passes through the point (2, 0), we have $-x_0^4=4x_0^3(2-x_0)$, or $3x_0^4-8x_0^3=0$
That is,$x_0=0$ or $x_0=8/3$, so that the points of tangency are (0, 0) and (8/3, 4096/81). Therefore, the equations of the tangents are
$y=0$ and $y-\frac{4098}{81}=\frac{2048}{27}(x-\frac{8}{3})$