Question

the point at which the tangent line to the curve x^3+y^3=a^3 is parallel to y axis​

Answer 1

Step-by-step explanation:

If a tangent at point (x1,y1) to a curve x^3+y^3=a^3 meets the curve again at point (x2,y2), how does one prove that x2/x1 + y2/y1 = - 1?

Given [math]x^3 + y^3 = a^3[/math]. The derivative is,

[math]\displaystyle \frac{dy}{dx} = -\frac{x^2}{y^2} \tag{1}[/math]

Therefore, slope of tangent at [math](x_1, y_1)[/math] is

[math]\displaystyle -\frac{x_1^2}{y_1^2} \tag{2}[/math]

The tangent passes through [math](x_2, y_2)[/math], therefore, slope of the tangent is also given by

[math]\displaystyle \frac{y_2 - y_1}{x_2 - x_1} \tag{3}[/math]

Comparing the two slope equations we get,

[math]\displaystyle \frac{y_2 - y_1}{x_2 - x_1} = -\frac{x_1^2}{y_1^2} \tag{4.1}[/math]

[math]\displaystyle \frac{y_2^3 - y_1^3}{x_2^3 - x_1^3} \times \frac{x_1^2 + x_1x_2 + x_2^2}{y_1^2 + y_1y_2 + y_2^2} = -\frac{x_1^2}{y_1^2} \tag{4.2}[/math]

[math]\displaystyle - \frac{x_1^2 + x_1x_2 + x_2^2}{y_1^2 + y_1y_2 + y_2^2} = -\frac{x_1^2}{y_1^2} \tag{4.3}[/math]

[math]\displaystyle x_1^2y_1^2 + x_1x_2y_1^2 + x_2^2y_1^2 = x_1^2y_1^2 + x_1^2y_1y_2 + x_1^2y_2^2 \tag{4.4}[/math]

[math]\displaystyle x_1x_2y_1^2 + x_2^2y_1^2 = x_1^2y_1y_2 + x_1^2y_2^2 \tag{4.5}[/math]

[math]\displaystyle x_1^2y_2^2 - x_2^2y_1^2 = x_1x_2y_1^2 - x_1^2y_1y_2 \tag{4.6}[/math]

[math]\displaystyle (x_1y_2 - x_2y_1)(x_1y_2 + x_2y_1) = x_1y_1(x_2y_1 - x_1y_2) \tag{4.7}[/math]

[math]\displaystyle x_1y_2 + x_2y_1 = -x_1y_1 \tag{4.8}[/math]

[math]\displaystyle \frac{x_2}{x_1} + \frac{y_2}{y_1} = -1 \tag{4.9}[/math]

Hope it helps

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