The length of the segment of the tangent line | vedclass

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(B) Given,the parametric equations of the curve are:$x = a \cos^3 t$$y = a \sin^3 t$Differentiating with respect to $t$:$\frac{dx}{dt} = -3a \cos^2 t

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$a$

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(B) Given,the parametric equations of the curve are:$x = a \cos^3 t$$y = a \sin^3 t$Differentiating with respect to $t$:$\frac{dx}{dt} = -3a \cos^2 t \sin t$$\frac{dy}{dt} = 3a \sin^2 t \cos t$The slope of the tangent is:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\sin t}{\cos t}$The equation of the tangent at $(a \cos^3 t, a \sin^3 t)$ is:$y - a \sin^3 t = -\frac{\sin t}{\cos t} (x - a \cos^3 t)$$y \cos t - a \sin^3 t \cos t = -x \sin t + a \cos^3 t \sin t$$x \sin t + y \cos t = a \sin t \cos t (\sin^2 t + \cos^2 t)$$x \sin t + y \cos t = a \sin t \cos t$To find the intercepts,set $y=0$ to get $x = a \cos t$,and set $x=0$ to get $y = a \sin t$.The points are $A(a \cos t, 0)$ and $B(0, a \sin t)$.The length of the segment $AB$ is:$L = \sqrt{(a \cos t - 0)^2 + (0 - a \sin t)^2} = \sqrt{a^2 \cos^2 t + a^2 \sin^2 t} = \sqrt{a^2(1)} = a$.

The length of the segment of the tangent line to the curve $x=a \cos ^3 t, y=a \sin ^3 t$,at any point on the curve cut off by the coordinate axes is

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