# 48 - 49 Shortest distance from a point to a curve by maxima and minima

**Problem 48**

Find the shortest distance from the point (5, 0) to the curve 2y^{2} = x^{3}.

**Solution:**

$d = \sqrt{(x - 5)^2 + y^2}$

from

$2y^2 = x^3$

$y^2 = \frac{1}{2} x^3$

$d = \sqrt{(x - 5)^2 + \frac{1}{2} x^3}$

$\dfrac{dd}{dx} = \dfrac{2(x - 5) + \frac{3}{2}x^2}{2\sqrt{(x - 5)^2 + \frac{1}{2} x^3}} = 0$

$\frac{3}{2}x^2 + 2x - 10 = 0$

$3x^2 + 4x - 20 = 0$

$(3x + 10)(x - 2) = 0$

For $3x + 10 = 0$, $x = -10/3$ (*meaningless*)

For $x - 2 = 0$, $x = 2$ (*okay*)

Use $x = 2$.

$d = \sqrt{(2 - 5)^2 + \frac{1}{2} (2^3)}$

$d = \sqrt{13}$ *answer*

**Another Solution:**

$4y \, y' = 3x^2$

$y' = \dfrac{3x^2}{4y}$ → *slope of tangent at any point*

Thus, the slope of normal at any point is

$m = -\dfrac{4y}{3x^2}$

Equation of normal:

$y - y_1 = m(x - x_1)$

$y - 0 = -\dfrac{4y}{3x^2}(x - 5)$

$3x^2 y = -4xy + 20y$

$3x^2 = -4x + 20$

$3x^2 + 4x - 20 = 0$ the same equation as above (*okay*)

**Problem 49**

Find the shortest distance from the point (0, 8a) to the curve ax^{2} = y^{3}.

**Solution:**

$d = \sqrt{x^2 + (y - 8a)^2}$

From

$ax^2 = y^3$

$x^2 = \dfrac{1}{a} y^3$

$d = \sqrt{\dfrac{1}{a} y^3 + (y - 8a)^2}$

$\dfrac{dd}{dy} = \dfrac{\dfrac{3}{a} y^2 + 2(y - 8a)}{2\sqrt{\dfrac{1}{a} y^3 + (y - 8a)^2}} = 0$

$\dfrac{3}{a} y^2 + 2y - 16a = 0$

$3y^2 + 2ay - 16a^2 = 0$

$y = \dfrac{-2a \pm \sqrt{4a^a - 4(3)(-16a^2)}}{2(3)}$

$y = \dfrac{-2a \pm 14a}{6}$

$y = 2a \, \text{ and } \, -\frac{8}{3}a$

$y = -\frac{8}{3}a$ is meaningless, use $y = 2a$

$d = \sqrt{\dfrac{1}{a} (2a)^3 + (2a - 8a)^2}$

$d = \sqrt{\dfrac{1}{a} (2a)^3 + (2a - 8a)^2}$

$d = \sqrt{44a^2}$

$d = 2a \sqrt{11}$ *answer*