I have no idea how to do this math problem.
x = (4(t^3) + 1)/4
y = 3(t^5) +1
Find all points (a,b) on this curve such that the tangent line to the curve at (a, b) passes through the point (1/5, 1).
Help?!
First we need some info on the point $\displaystyle (a,b)$ putting this into the equation for x gives
$\displaystyle \displaystyle a=t^3+\frac{1}{4} \iff t=\left(a-\frac{1}{4}\right)^{\frac{1}{3}}$
Now putting this into the equation for y gives
$\displaystyle \displaystyle y=3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1$
So the ordered pair $\displaystyle \displaystyle \left(a,3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1\right)$ must lie on the graph.
Now we can use this generic point to find the slope passing through the point $\displaystyle \displaystyle \left( \frac{1}{5},1\right)$ this gives
$\displaystyle \displaystyle m=\frac{3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}+1-1}{a-\frac{1}{5}}= \frac{3\left(a-\frac{1}{4}\right)^{\frac{5}{3}}}{a-\frac{1}{5}}$
Now we also know the slope at this point must be
$\displaystyle \displaystyle \frac{dy}{dx}\bigg|_{x=a}$ so...
Can you finish from here?
$\displaystyle \displaystyle x = {{4\,t^3 + 1}\over4}\ \ \to\ \ x = t^3 + {1\over4}$
$\displaystyle y = 3\,t^5 +1$
$\displaystyle \displaystyle {{dy}\over{dx}}={{dy/dt}\over{dx/dt}}={{15t^4}\over{3t^2}} =5t^2$
Slope from $\displaystyle \displaystyle ({1\over5},\ 1)$ to $\displaystyle (x,\ y)$ must equal $\displaystyle 5t^2$.
$\displaystyle \displaystyle {{3\,t^5 +1-1}\over{t^3 + {1\over4}-{1\over5}}} =5t^2$.
Solve for $\displaystyle \displaystyle t$ (There are 2 solutions.) and use $\displaystyle \displaystyle t$ to find $\displaystyle (x,\ y)$.