I have parabola equation (notice it's "sideways", given x =)
$\displaystyle x = \frac{y^2}{2}-4y+3$ and a point $(2, -1)$.
Find parabola tangents (equations) that go through point $(2,-1)$.
Ive watched several videos and looked similar solved problems, however I'm getting it wrong somewhere. Maybe it's because I cant take into account that parabola is given through $x$ and not $y$. Im not sure. But here is what I tried to do.
I take derivative of parabola equation : $x'= y-4$ so $x'$ is the slope of tangents
I can take some point on parabola $x,y$ and replace $x$ with equation so point is $(\frac{y^2}{2}-4y+3, y)$ now I put that point into slope forumla with given point
$\displaystyle \frac{y-(-1)} {\frac{y^2}{2}-4y+3} - 2 = y - 4$
if I slove this I should get y coordinates of points where tangents touch parabola ? However the results do not make sense and the resulting formulas I make for tangents are not correct.
There is correct answer given to this problem, which is tangents equations are either
y = −0,1202...x − .0, 7595..
or
y = −0,5940...x + 0,188
I cant reach them, what am I missing here?
Your derivative is incorrect and hence you are getting wrong answer.
Parabola equation is $x = \frac{y^2}{2}-4y+3$
you should differentiate wrt. $x$
$1 = y \frac{dy}{dx} - 4 \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{y-4}$
So equation of tangent line is
$y + 1 = \frac{1}{y_0-4} (x -2)$ ...(i)
(if the tangent goes through point $(x_0, y_0)$ on the parabola).
As the point $(x_0, y_0)$ is both on the line and the parabola,
$y_0 + 1 = \frac{1}{y_0-4} (x_0 -2) \implies y_0^2 - 3y_0 -2 = x_0$
and also $x_0 = \frac{y_0^2}{2}-4y_0+3$
Solving both,
$y_0 = \pm \sqrt 11 - 1$ gives you slopes $(\frac{1}{y_0-4}) \approx -0.59404, -0.12024$
Substituting both values in equation $(i)$, you get two tangents to parabola passing through point $(2, -1)$.