Question

Find the equations of the tangents drawn to the curve $y=x^4$ which are drawn from the point (2,0).

• A. $y=0$ and $y-{\left(\frac{4}{3}\right)}^4=4{\left(\frac{4}{3}\right)}^3(x-\frac{4}{3})$
• B. $y=0$ and $y-{\left(\frac{8}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^3(x-\frac{8}{3})$
• C. $y=0$ and $y-{\left(\frac{4}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^2(x-\frac{8}{3})$
• D. $y=0$ and $y-{\left(\frac{8}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^2(x-\frac{8}{3})$

Answer 1

Answer: (B)

$y=0$ and $y-{\left(\frac{8}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^3(x-\frac{8}{3})$

Let the equation of curve be $y=x^4$
Let the point of contact be $(h,k)$
$\Rightarrow k=h^4$ ...(1)
Now, $\frac{dy}{dx}=4x^3$
Slope of tangent to the curve at $(h,k)$ is $4h^3$
Tangent is $y-k=4h^3(x-h)$ ...(2)
It passes through (2,0)
$\therefore -k=4h^3(2-h)$
$-h^4=8h^3-4h^4$ by (1)
$3h^4-8h^3=0$
$\therefore h=0$ or $\frac{8}{3}$
$\therefore k=0$ or ${\left(8/3\right)}^4$
$\therefore$ Points are (0,0) and $[8/3,{\left(8/3\right)}^4]$
Putting in (2), tangents are $y=0$
and $y-{\left(\frac{8}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^3(x-\frac{8}{3})$