The equation of a tangent to the parabola,x^2 | vedclass

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Question

(C) Given the parabola equation is $x^2 = 8y$.Differentiating with respect to $x$,we get $2x = 8 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{x

Vedclass · Accepted Answer

$x = y \cot 	heta + 2 	an 	heta$

Vedclass · Answer

(C) Given the parabola equation is $x^2 = 8y$.Differentiating with respect to $x$,we get $2x = 8 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{x}{4}$.Since the tangent makes an angle $	heta$ with the positive $x-$axis,the slope of the tangent is $	an 	heta$.Thus,$\frac{x}{4} = 	an 	heta$,which gives $x = 4 	an 	heta$.Substituting $x = 4 	an 	heta$ into the parabola equation $x^2 = 8y$,we get $(4 	an 	heta)^2 = 8y$,so $16 	an^2 	heta = 8y$,which gives $y = 2 	an^2 	heta$.The point of contact is $(4 	an 	heta, 2 	an^2 	heta)$.The equation of the tangent at $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$,where $m = 	an 	heta$.$y - 2 	an^2 	heta = 	an 	heta (x - 4 	an 	heta)$$y - 2 	an^2 	heta = x 	an 	heta - 4 	an^2 	heta$$y = x 	an 	heta - 2 	an^2 	heta$.Alternatively,dividing by $	an 	heta$ (assuming $	an 	heta 
eq 0$):$x = y \cot 	heta + 2 	an 	heta$.

The equation of a tangent to the parabola,$x^2 = 8y,$ which makes an angle $\theta$ with the positive direction of the $x-$axis,is

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