If the tangent at any point on the curve x^4 | vedclass

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Question

(A) Given the curve $x^4 + y^4 = a^4$.Let the point of tangency be $P(x_1, y_1)$.Differentiating with respect to $x$,we get $4x^3 + 4y^3 \frac{dy}{dx}

Vedclass · Accepted Answer

$a^{-4/3}$

Vedclass · Answer

(A) Given the curve $x^4 + y^4 = a^4$.Let the point of tangency be $P(x_1, y_1)$.Differentiating with respect to $x$,we get $4x^3 + 4y^3 \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x^3}{y^3}$.The slope of the tangent at $(x_1, y_1)$ is $m = -\frac{x_1^3}{y_1^3}$.The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = -\frac{x_1^3}{y_1^3}(x - x_1)$.This simplifies to $y y_1^3 - y_1^4 = -x x_1^3 + x_1^4$,or $x x_1^3 + y y_1^3 = x_1^4 + y_1^4$.Since $(x_1, y_1)$ lies on the curve,$x_1^4 + y_1^4 = a^4$,so the equation is $x x_1^3 + y y_1^3 = a^4$.The $x$-intercept $p$ is found by setting $y=0$,so $p = \frac{a^4}{x_1^3}$.The $y$-intercept $q$ is found by setting $x=0$,so $q = \frac{a^4}{y_1^3}$.Now,calculate $p^{-4/3} + q^{-4/3} = (\frac{a^4}{x_1^3})^{-4/3} + (\frac{a^4}{y_1^3})^{-4/3}$.$= a^{-16/3} (x_1^4 + y_1^4) = a^{-16/3} (a^4) = a^{4 - 16/3} = a^{-4/3}$.

If the tangent at any point on the curve $x^4 + y^4 = a^4$ meets the axes at $p$ and $q$,then the value of $p^{-4/3} + q^{-4/3}$ is:

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