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

$d = \sqrt{(x - 5)^2 + y^2}$
 

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

$y' = \dfrac{3x^2}{4y}$   →   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)

$d = \sqrt{x^2 + (y - 8a)^2}$
 

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