# A parabola has a critical point at (25, -14). It also has a tangent with equation y=-18x+20 . What is the equation of the parabola?

Apr 26, 2017

$y = \frac{81}{416} {x}^{2} - \frac{2025}{208} x + \frac{44801}{416}$

#### Explanation:

Suppose the required parabola has the equation:

$y = a {x}^{2} + b x + c$

We differentiate wrt $x$ to get the first derivative:

$y ' = 2 a x + b$

We want a critical point at $\left(25 , - 14\right)$, so we can use $y ' = 0$ at that point:

$x = 25 \implies 2 \cdot 25 a + b = 0$

$\therefore 50 a + b = 0 \implies b = - 50 a$

This critical point $\left(25 , - 14\right)$ also lies on the original curve:

$x = 25 \implies 625 a + 25 b + c = - 14$

$\therefore 625 a + 25 \left(- 50 a\right) + c = - 14$
$\therefore 625 a - 1250 a + c = - 14$
$\therefore 625 a - c = 14 \implies c = 625 a - 14$

We also require one simultaneous solution of:

$y = a {x}^{2} + b x + c$
$y = - 18 x + 20$

So that:

$a {x}^{2} + b x + c = - 18 x + 20$
$\therefore a {x}^{2} + \left(b + 18\right) x + c - 20 = 0$

For one solution then the discriminant must be zero:

$\therefore {\left(b + 18\right)}^{2} - 4 \left(a\right) \left(c - 20\right) = 0$

Substituting $b = - 50 a$ and $c = 625 a - 14$ gives:

$\therefore {\left(- 50 a + 18\right)}^{2} - 4 a \left(625 a - 14 - 20\right) = 0$
$\therefore 2500 {a}^{2} - 1800 a + 324 - 2500 {a}^{2} + 136 a = 0$
$\therefore - 1664 a + 324 = 0$
$\therefore a = \frac{81}{416}$

And so:

$b = - \frac{2025}{208}$
$c = \frac{44801}{416}$

Hence the equation we seek is

$y = \frac{81}{416} {x}^{2} - \frac{2025}{208} x + \frac{44801}{416}$

We can validate the solution graphically: