Question

Find an equation of the line tangent to the curve y=3x²-1 and perpendicular to the line x-3y=4

Answer 1

Let y=g(x) be a sought line.

The equation of y=g(x) could be found the following way: $g(x) - g{\scriptscriptstyle 0} = k{\scriptscriptstyle 1}(x-x{\scriptscriptstyle 0})$, where ($x{\scriptscriptstyle 0},y{\scriptscriptstyle 0})$ is the point belonging to a line, $k{\scriptscriptstyle 1}$ is the slope of the line.

Since y=g(x) is perpendicular to x-3y=4, then $k{\scriptscriptstyle 1}k{\scriptscriptstyle 2}=-1$, where $k{\scriptscriptstyle 2}$ is the slope of the second line.

$x-3y=4\to 3y=x-4\to y = {\frac x 3} - {\frac 4 3} \to k{\scriptscriptstyle 2} = {\frac 1 3}$

${\frac 1 3}k{\scriptscriptstyle 1} = -1 \to k{\scriptscriptstyle 1} = -3$

g(x) is tangent to $y = 3x^{2} -1$. The slope for the tangent line to $y = 3x^{2} -1$ is $y'=6x$

Then $6x_0 = -3$

$x{\scriptscriptstyle 0}$ = -0.5

$y{\scriptscriptstyle 0} = 3*(-0.5)^{2} -1 = -0.25$

Then $g(x) + 0.25 = -3(x +0.5) \to g(x) = -3x - 1.75$

So, the equation of a sought line is $4y + 12x + 7$ = 0