Find multiple tangents for parabola

Question

I have a parabola $x = 0.5y^2 - 4y + 3$. I need to find a tangent(s) that go throught point $(2,-1)$. I have tried:

$y - (-1) = m(x-2)$

$4y = 0.5y^2 - x + 3$

$y = (0.5y^2)/4 - x/4 + 3/4$

$m(x-2) - 1 = (0.5y^2)/4 - x/4 + 3/4$

But I'm pretty sure that's incorrect, could I have some advise on the problem? Thanks a lot

Answer 1

Equation of parabola is $x = 0.5y^2 - 4y + 3$. You can write tangent points to the parabola as $(0.5t^2 - 4t + 3, t)$.

Slope at the tangent point $ = \displaystyle \frac{dy}{dx} = \frac{1}{t-4}$

So the tangent lines are given by,

$\displaystyle y-t = \frac{1}{t-4} (x - 0.5t^2 + 4t - 3) $

As the tangents go through point $(2, -1)$

$\displaystyle -1-t = \frac{1}{t-4} (2 - 0.5t^2 + 4t - 3) $

Solving, $t = -1 \pm \sqrt{11}$

So equation of tangent lines can be written as,

$ \displaystyle y + 1 = \frac{1}{-5 \pm \sqrt{11}} (x - 2)$

Answer 2

Can you see where these equations come from?

$$ \begin{cases} x= 2 + m(y+1) &\qquad&(1)\\ m = -4+y &&(2)\\ x = \frac 12 y^2 -4y +3 &&(3)\\ \end{cases} $$

The tangency point belongs to both the parabola and the tangent line (equations 1 and 3), and the slope of the tangent line is the derivative at the tangency point (equation 2).

This way you find two solutions, for $y=-1\pm\sqrt{11}$, with slopes $-5 \pm \sqrt{11}$.