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?

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Answer 1 · 2017-04-26T01:52:48

$y=81/416x^2 -2025/208x+44801/416$

Suppose the required parabola has the equation:

$y=ax^2+bx+c$

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

$y'=2ax+b$

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

$x=25 => 2*25a+b = 0$

$:. 50a+b = 0 => b=-50a$

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

$x=25 => 625a+25b+c=-14$

$:. 625a+25(-50a)+c=-14$
$:. 625a-1250a+c=-14$
$:. 625a-c=14 => c=625a-14$

We also require one simultaneous solution of:

$y=ax^2+bx+c$
$y=-18x+20$

So that:

$ax^2+bx+c = -18x+20$
$:. ax^2+(b+18)x+c-20 = 0$

For one solution then the discriminant must be zero:

$:. (b+18)^2 - 4(a)(c-20) = 0$

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

$:. (-50a+18)^2 - 4a(625a-14-20) = 0$
$:. 2500a^2 -1800a+324 -2500a^2+136a = 0$
$:. -1664a+324 = 0$
$:. a=81/416$

And so:

$b=-2025/208$
$c= 44801/416$

Hence the equation we seek is

$y=81/416x^2 -2025/208x+44801/416$

We can validate the solution graphically:
enter image source here

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