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
$x = 0$ ← the y-axis
The equation of normal is therefore the x-axis $y = 0$ answer
At (1, 0), x1 = 1 and y1 = 0
$x - y - \frac{1}{2}x - \frac{1}{2} = 0$
$\frac{1}{2}x - y - \frac{1}{2} = 0$
$y = \frac{1}{2}x - \frac{1}{2}$ ← 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)
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)
More information about text formats
Follow @iMATHalino
MATHalino
$x^2 + 2y^2 - 2xy - x = 0$
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
$x = 0$ ← the y-axis
The equation of normal is therefore the x-axis
$y = 0$ answer
At (1, 0), x1 = 1 and y1 = 0
$x - y - \frac{1}{2}x - \frac{1}{2} = 0$
$\frac{1}{2}x - y - \frac{1}{2} = 0$
$y = \frac{1}{2}x - \frac{1}{2}$ ← 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)
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)
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
Add new comment