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 .

- Tapered Beam
- Vickers hardness: Distance between indentations
- Time rates
- Minima Maxima: y=ax³+bx²+cx+d
- Make the curve y=ax³+bx²+cx+d have a critical point at (0,-2) and also be a tangent to the line 3x+y+3=0 at (-1,0).
- Minima maxima: Arbitrary constants for a cubic
- Minima Maxima: 9a³y=x(4a-x)³
- Minima maxima: a²y = x⁴
- how to find the distance when calculating moment of force
- strength of materials
- Analytic Geometry Problem Set [Locked: Multiple Questions]
- Equation of circle tangent to two lines and passing through a point
- Product of Areas of Three Dissimilar Right Triangles
- Perimeter of Right Triangle by Tangents
- Differential equations
- Laplace
- Families of Curves: family of circles with center on the line y= -x and passing through the origin
- Family of Plane Curves
- Differential equation
- Differential equation

Home • Forums • Blogs • Glossary • Recent

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

About • Contact us • Disclaimer • Privacy Policy • Hosted by WebFaction • Powered by Drupal

Forum posts (unless otherwise specified) licensed under a Creative Commons Licence.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

All trademarks and copyrights on this page are owned by their respective owners. Forum posts are owned by the individual posters.

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

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

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

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

$y = -2x + 2$

$y + 2x - 2 = 0$

answer