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 (x_{1}, y_{1}) $xx_1 + 2yy_1 - 2\left( \dfrac{xy_1 + yx_1}{2} \right) - \left( \dfrac{x + x_1}{2} \right) = 0$

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

At (0, 0), x_{1} = 0 and y_{1} = 0

$x = 0$ ← the y-axis

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

At (1, 0), x_{1} = 1 and y_{1} = 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)

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$x^2 + 2y^2 - 2xy - x = 0$

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 (x

_{1}, y_{1})$xx_1 + 2yy_1 - 2\left( \dfrac{xy_1 + yx_1}{2} \right) - \left( \dfrac{x + x_1}{2} \right) = 0$

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

At (0, 0), x

_{1}= 0 and y_{1}= 0$x = 0$ ← the y-axis

The equation of normal is therefore the x-axis

$y = 0$

answerAt (1, 0), x

_{1}= 1 and y_{1}= 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

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

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

$y = -2x + 2$

$y + 2x - 2 = 0$

answerYou 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

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

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

$y = 0$

answerAt (1, 0)

For normal line, the slope m = -2

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

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

$y = -2x + 2$

$y + 2x - 2 = 0$

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