Question

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient. Find the equation of L in the form y = mx + c.$y= 4x^{2} +1$

Answer 1

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$, so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.