The x-intercept of the tangent at any arbitra | vedclass

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(C) Given the curve equation: $\frac{a}{x^2} + \frac{b}{y^2} = 1$. Differentiating both sides with respect to $x$: $-\frac{2a}{x^3} - \frac{2b}{y^3} \

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cube of the abscissa of the point of tangency

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(C) Given the curve equation: $\frac{a}{x^2} + \frac{b}{y^2} = 1$. Differentiating both sides with respect to $x$: $-\frac{2a}{x^3} - \frac{2b}{y^3} \frac{dy}{dx} = 0$. This gives the slope of the tangent: $\frac{dy}{dx} = -\frac{ay^3}{bx^3}$. The equation of the tangent at point $(x, y)$ is: $Y - y = -\frac{ay^3}{bx^3}(X - x)$. To find the $x-$intercept,set $Y = 0$: $-y = -\frac{ay^3}{bx^3}(X - x) \implies X - x = \frac{bx^3}{ay^2} \implies X = x + \frac{bx^3}{ay^2}$. Simplifying the expression: $X = x \left( 1 + \frac{bx^2}{ay^2} \right) = x \left( \frac{ay^2 + bx^2}{ay^2} \right)$. Since $\frac{a}{x^2} + \frac{b}{y^2} = 1$,we have $ay^2 + bx^2 = x^2y^2$. Substituting this into the expression for $X$: $X = x \left( \frac{x^2y^2}{ay^2} \right) = \frac{x^3}{a}$. Thus,the $x-$intercept is proportional to the cube of the abscissa $x^3$.

The $x-$intercept of the tangent at any arbitrary point of the curve $\frac{a}{x^2} + \frac{b}{y^2} = 1$ is proportional to

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