The area of the triangle enclosed by the axes and the tangent to the curve y = 1/ x , at any point on the parabola is
Now use the formula $$yy_o=m(xx_o)$$ $$yy_o=\frac{1}{x_o^2}(xx_o)$$ $$y\frac{1}{x_o}=\frac{1}{x_o^2}(xx_o)$$ $$0\frac{1}{x_o}=\frac{1}{x_o^2}(xx_o)$$ Now multiply both sides of the equation with $x_o^2$ $$x_o^2\left(\frac{1}{x_o}\right)=\frac{1}{x_o^2}(xx_o)\cdot x_o^2$$ $$x_o=1(xx_o)$$ $$2x_o=x$$ This is value of $x$ intercept and in the similar way find the value of $y$ intercept. $$yy_o=m(xx_o)$$ $$yy_o=\frac{1}{x_o^2}(xx_o)$$ $$y\frac{1}{x_o}=\frac{1}{x_o^2}(xx_o)$$ $$y\frac{1}{x_o}=\frac{1}{x_o^2}(0x_o)$$ $$y\frac{1}{x_o}=\frac{x_o}{x_o^2}$$ $$y\frac{1}{x_o}=\frac{1}{x_o}$$ $$y=\frac{2}{x_o}$$
Now use then use the formula $\dfrac12\cdot$base $\cdot$height
$$=\frac12\cdot 2x_o\cdot \frac{2}{x_o}=2$$
So, the area of the triangle enclosed by the axes and the tangent to the parabola $y=\dfrac1x$ at any point on the parabola is $2$ sq.units
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Comments

nash about 1 year
Consider the following diagram: A point on the curve is chosen, and the tangent line is drawn until it intersects the axes, delimiting a triangle (in grey). How can we find the area of the triangle, for an arbitrary chosen point on the curve?
I can't figure out what to do after finding the slope of tangent

Crosby over 5 yearsPlease consider using mathjax to format your answer.

Admin over 5 yearsNice, where did you draw it?