How do you calculate the gradient of tangent to xy=4 and x^2=16y?

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How do you calculate the gradient of tangent to xy=4 and x^2=16y?

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Answer 1 · 2018-07-11T08:32:30

At a point $(x_{1}, y_{1})$ $(x_1,y_1)$ , the gradient is $- \frac{4}{x_{1}^{2}}$ $-4/x_1^2$ , when it is on curve $x y = 4$ $xy=4$ and $\frac{x_{1}}{8}$ $x_1/8$ , when it is on $x^{2} = 16 y$ $x^2=16y$ .

Explanation:

Gradient of a curve is given by the slope of curve at that point, whose value is value of the derivative at that point.

(1) Here $x y = 4$ $xy=4$ i.e. $y = \frac{4}{x}$ $y=4/x$ and $\frac{d y}{d x} = - \frac{4}{x^{2}}$ $(dy)/(dx)=-4/x^2$ .

Now at point $(x_{1}, y_{1})$ $(x_1,y_1)$ on the curve (observe $x_{1} y_{1} = 4$ $x_1y_1=4$ ),

the value of derivative of gradient of curve is $- \frac{4}{x_{1}^{2}}$ $-4/x_1^2$

(1) Here $x^{2} = 16 y$ $x^2=16y$ or $y = \frac{x^{2}}{16}$ $y=x^2/16$ and $\frac{d y}{d x} = \frac{x}{8}$ $(dy)/(dx)=x/8$ .

Now at point $(x_{1}, y_{1})$ $(x_1,y_1)$ on the curve (observe $x_{1}^{2} = 16 y_{1}$ $x_1^2=16y_1$ ),

the value of derivative of gradient of curve is $\frac{x_{1}}{8}$ $x_1/8$

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How do you calculate the gradient of tangent to xy=4 and x^2=16y?

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