# 50 - 52 Nearest distance from a given point to a given curve

**Problem 50**

Find the shortest distance from the point (4, 2) to the ellipse x^{2} + 3y^{2} = 12.

**Solution:**

from

$x^2 + 3y^2 = 12$

$x = \sqrt{12 - 3y^2}$

$d = \sqrt{\left( \sqrt{12 - 3y^2} - 4 \right)^2 + (y - 2)^2}$

$\dfrac{dd}{dy} = \dfrac{2\left( \sqrt{12 - 3y^2} - 4 \right) \dfrac{-6y}{2\sqrt{12 - 3y^2}} + 2(y - 2)}{2\sqrt{\left( \sqrt{12 - 3y^2} - 4 \right)^2 + (y - 2)^2}} = 0$

$-6y + \dfrac{24y}{\sqrt{12 - 3y^2}} + 2y - 4 = 0$

$\dfrac{24y}{\sqrt{12 - 3y^2}} = 4y + 4$

$\dfrac{6y}{\sqrt{12 - 3y^2}} = y + 1$

$\dfrac{36y^2}{12 - 3y^2} = (y + 1)^2$

$\dfrac{36y^2}{3(4 - y^2)} = (y + 1)^2$

$12y^2 = (y + 1)^2 (4 - y^2)$

$12y^2 = (y^2 + 2y + 1) (4 - y^2)$

$12y^2 = 4y^2 + 8y + 4 - y^4 - 2y^3 - y^2$

$y^4 + 2y^3 + 9y^2 - 8y - 4 = 0$

By trial and error

$y = 1$

$x = \sqrt{12 - 3(1^2)} = 3$

The nearest point is (3, 1)

Nearest distance:

$d = \sqrt{(3 - 4)^2 + (1 - 2)^2}$

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

**Another Solution:**

$x^2 = 12 - 3y^2$

$2x = -6y \, y'$

$y' = -\dfrac{x}{3y}$ → slope of tangent at any point

Thus, slope of normal at any point is

$m = \dfrac{3y}{x}$

Equation of normal:

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

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

$xy - 2x = 3xy - 12y$

$x + xy - 6y = 0$

$x(1 + y) = 6y$

$x^2 (1 + y)^2 = 36y^2$

$(12 - 3y^2)(1 + 2y + y^2) = 36y^2$

$12 + 24y + 12y^2 - 3y^2 - 6y^3 - 3y^4 = 36y^2$

$3y^4 + 6y^3 - 27y^2 - 24y - 12 = 0$

$y^4 + 2y^3 + 9y^2 - 8y - 4 = 0$ the same equation as above (*okay*)

**Problem 51**

Find the shortest distance from the point (1 + n, 0) to the curve y = x^{n}, n > 0.

**Solution:**

$d = \sqrt{[ \, x - (1 + n) \, ]^2 + x^{2n}}$

$\dfrac{dd}{dx} = \dfrac{2[ \, x - (1 + n) \, ] + 2nx^{2n - 1}}{2\sqrt{[ \, x - (1 + n) \, ]^2 + x^{2n}}} = 0$

$x - 1 - n + nx^{2n - 1} = 0$

$x + nx^{2n - 1} = 1 + n$

by inspection: x = 1

$d = \sqrt{[ \, 1 - (1 + n) \, ]^2 + 1^{2n}}$

$d = \sqrt{n^2 + 1^{2n}}$ 1 raise to any positive number is 1

$d = \sqrt{1 + n^2}$ *answer*

**Problem 52**

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

**Solution:**

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

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

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

Thus, slope of normal at any point is

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

Equation of normal:

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

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

$xy - 5x = -2y$

$(x + 2)y = 5x$

$(x + 2)^2 \, y^2 = 25x^2$

$(x^2 + 4x + 4)(\frac{1}{3}x^3) = 25x^2$

$x^3 + 4x^2 + 4x - 75 = 0$

By trial and error

$x = 3$

$y^2 = \frac{1}{3}(3^3) = 3^2$

$y = 3$

Nearest point on the curve is (3, 3)

Shortest distance

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

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

For similar problem, see Problem 48 and its solution.