46 - 47 Solved Problems in Maxima and Minima

For equilateral hyperbola, b = a.
$\dfrac{x^2}{a^2} - \dfrac{y^2}{a^2} = 1$
 

Thus,
$x^2 - y^2 = a^2$

$x^2 = a^2 + y^2$
 

Distance d:
$d = \sqrt{(x - 0)^2 + (y - k)^2}$

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

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

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

$2y + 2(y - k)$

$2y = k$

$y = \frac{1}{2} k$

$y^2 = \frac{1}{4} k^2$
 

Nearest Distance:
$d = \sqrt{a^2 + \frac{1}{4} k^2 + (\frac{1}{2} k - k)^2}$

$d = \sqrt{a^2 + \frac{1}{2} k^2}$           answer

$d = \sqrt{(x - 4a)^2 + y^2}$
 

from
$a^2 y = x^3$

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

$y^2 = \dfrac{x^6}{a^4}$
 

$d = \sqrt{(x - 4a)^2 + \dfrac{x^6}{a^4}}$

$\dfrac{dd}{dx} = \dfrac{2(x - 4a) + \dfrac{6x^5}{a^4}}{2\sqrt{(x - 4a)^2 + \dfrac{x^6}{a^4}}} = 0$

$(x - 4a) + \dfrac{3x^5}{a^4} = 0$

$3x^5 + a^4 x - 4a^5 = 0$
 

by trial and error:
$x = a$

$y = \dfrac{a^3}{a^2} = a$
 

The nearest point is (a, a).           answer