Equation of Line: Normal to conic with product xy in the equation

 
 

Find the equation of the normal line to the curve x^2 + 2y^2 - 2xy - x = 0 at the points where it crosses the x-axis .

$x^2 + 2y^2 - 2xy - x = 0$
 

At the points where it crosses the x-axis, y = 0
$x^2 - x = 0$

$x(x - 1) = 0$

$x = 0 ~ \text{and} ~ 1$   ←   hence, x-intercepts are (0, 0) and (1, 0)
 

Equation of tangent at any point (x1, y1)
$xx_1 + 2yy_1 - 2\left( \dfrac{xy_1 + yx_1}{2} \right) - \left( \dfrac{x + x_1}{2} \right) = 0$

$xx_1 + 2yy_1 - xy_1 - yx_1 - \frac{1}{2}x - \frac{1}{2}x_1 = 0$
 

At (0, 0), x1 = 0 and y1 = 0

$0 + 2(0) - 0 - 0 - \frac{1}{2}x - 0 = 0$

$x = 0$   ←   the y-axis
 

The equation of normal is therefore the x-axis
$y = 0$           answer

 

At (1, 0), x1 = 1 and y1 = 0

$x + 2(0) - 0 - y - \frac{1}{2}x - \frac{1}{2} = 0$

$x - y - \frac{1}{2}x - \frac{1}{2} = 0$

$\frac{1}{2}x - y - \frac{1}{2} = 0$

$y = \frac{1}{2}x - \frac{1}{2}$   ←   equation of tangent
 

For normal line, the slope m = -2
$y - y_1 = m(x - x_1)$

$y - 0 = -2(x - 1)$

$y = -2x + 2$

$y + 2x - 2 = 0$           answer

 
You can also use Calculus to find the slopes of the tangent:
$x^2 + 2y^2 - 2xy - x = 0$

$2x + 4y \, y' - 2(xy' + y) - 1 = 0$

$2x + (4y - 2x) \, y' - 2y - 1 = 0$

$y' = \dfrac{1 + 2y - 2x}{4y - 2x}$
 

At (0, 0)

$y' = \infty$   ←   vertical line
 

For normal line, the slope is horizontal line or m = 0
$y - y_1 = m(x - x_1)$

$y - 0 = 0(x - 0)$

$y = 0$           answer

 

At (1, 0)

$y' = \dfrac{1 + 0 - 2}{0 - 2} = \dfrac{1}{2}$
 

For normal line, the slope m = -2
$y - y_1 = m(x - x_1)$

$y - 0 = -2(x - 1)$

$y = -2x + 2$

$y + 2x - 2 = 0$           answer

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