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

calculus

1,891

Now use the formula $$y-y_o=m(x-x_o)$$ $$y-y_o=\frac{-1}{x_o^2}(x-x_o)$$ $$y-\frac{1}{x_o}=\frac{-1}{x_o^2}(x-x_o)$$ $$0-\frac{1}{x_o}=\frac{-1}{x_o^2}(x-x_o)$$ Now multiply both sides of the equation with $x_o^2$ $$x_o^2\left(-\frac{1}{x_o}\right)=\frac{-1}{x_o^2}(x-x_o)\cdot x_o^2$$ $$-x_o=-1(x-x_o)$$ $$2x_o=x$$ This is value of $x$ intercept and in the similar way find the value of $y$ intercept. $$y-y_o=m(x-x_o)$$ $$y-y_o=\frac{-1}{x_o^2}(x-x_o)$$ $$y-\frac{1}{x_o}=\frac{-1}{x_o^2}(x-x_o)$$ $$y-\frac{1}{x_o}=\frac{-1}{x_o^2}(0-x_o)$$ $$y-\frac{1}{x_o}=\frac{x_o}{x_o^2}$$ $$y-\frac{1}{x_o}=\frac{1}{x_o}$$ $$y=\frac{2}{x_o}$$

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

1,891

nash

Fresh CSE Grad , i mainly code in java and python :)

Updated on August 20, 2022

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 years

Please consider using mathjax to format your answer.
Admin over 5 years

Nice, where did you draw it?