CalcGeo: A short story about areas.

First thing, the slope of the tangent line is the derivative evaluated at the point of intersection. We know that: $\dfrac{\mathrm d}{\mathrm dx}\dfrac1x=-\dfrac{1}{x^2}.$ So the slope of the tangent line at the point $x_0$ is: $\text{Slope of the tangent line at }x_0=-\dfrac{1}{x_0^2}.$ From this, we can find the $x$ -intercept by setting $y=0$ and solving using the slope-point formula: $\begin{aligned} y-y_0=-\dfrac{1}{x_0^2}(x-x_0)\xrightarrow[y=0]{}&0-y_0=-\dfrac{1}{x_0^2}(x-x_0)\\ &\Rightarrow -\dfrac1{x_0}=-\dfrac{1}{x_0^2}(x-x_0)=-\dfrac x{x_0^2}+\dfrac{x_0}{x_0^2}\\ &\Rightarrow -\dfrac1{x_0}=-\dfrac x{x_0^2}+\dfrac{1}{x_0}\\ &\Rightarrow \dfrac x{x_0^2}=\dfrac2{x_0}\Rightarrow \boxed{x=2x_0}. \end{aligned}$ So the point of intersection with the $x$ -axis is $x=2x_0$ , by symmetry, the $y$ -intercept is $y=2y_0$ since $\begin{aligned} y=\dfrac1x&\iff xy=1\\ &\iff x=\dfrac1y. \end{aligned}$ The area of a triangle is $\dfrac12hb$ , so the area of our triangles is: $\dfrac122x_0\cdot2y_0=2x_0y_0=2x_0\cdot\dfrac1{x_0}=\boxed{2}.$ See the image here to understand some of the notation: http://i.stack.imgur.com/06y57.png

By co-ordinate geometry, the vertex is (1,1). The asymptote being $y=x$ and the slope of the tangent touching the vertex is -1. So the tangent equation becomes $y=2 - x$ where x and y intercepts are (2,0) and (0,2) respectively. Hence the area is $\frac 12 \times 2 \times 2= 2 .\square$

$y'=-\frac{1}{x^2}$ at point (1,1), the tangent line has equation: y=-x+2, this line cut X axis at (2,0) $area=\int_0^2 dx \int_0^{-x+2} dy=\boxed{2}$

We can think about this problem starting with two functions: y = 1/x and y = mx + b (this comes We can think about this problem starting with two functions: y = 1/x and y = mx + b (this comes from the tangent line). We can then substitute y with 1/x in the second equation, giving us 1/x = mx + b. We can now turn this into a quadratic equation and set it equal to zero. $x (1/x ) = x ( mx + b )$ $1 = mx^2 + bx$ $mx^{ 2 } + bx + 1 = 0$

Now, lets say that the tangent line's y-intercept is b and its x-intercept is a. The b in y=mx+b is now the same as this b. This gives us the slope -b/a that we can now plug into the quadratic equation
$-b/a x^{ 2 } + bx + 1$

We can now plug this into the quadratic formula.
$(-b ± √b^{ 2 } - 4[-b/a][1])/2$

However, since the function I only touches the tangent line ONCE, there can only be one answer. Because this is a quadratic, in order to have only one answer, the discriminant must equal zero. So we set the discriminant equal to zero.

$b^{ 2 } - 4[-b/a] = 0$ $b^{ 2 } = 4b/a$ $b = 4/a$ $ba = 4$

Because the area of a triangle is 1/2 bh (1/2ba for this triangle), the answer is simply half of the above equation.

$\frac{1}{2} ba = 2$

The area of the triangle is 2.

A = (1,1) We write the beam of the straight line that switches for A and is tangent to y=1/x.. y-1=m(x-1) 1/x -1 = mx - m 1-x=mx^2-mx mx^2-x(m-1)-1=0 Δ=0 m^2+2m+1=0 (m+1)=0 m=-1 We replace in the beam and we have y=-x+2. So the others points are B=(0,2) C=(2,0) . We've have a triangle with 2 cathetus of 2 unities. So area is 2x2/2= 2

y=1/x => slope at any point, dy/dx = -1/x² $$ $$ at any point (x₀, y₀), $$ dy/dx = -1/x₀² $$ $$ => tangent line is y = (-1/x₀²)x + c $$ $$ since the tangent line satisfies by (x₀, y₀) $$ => c = y₀ + (1/x₀) $$ $$ so, the equation is, y = (-1/x₀²)x + y₀ + (1/x₀) $$ $$ now find x and y axis cut point, $$ when x = 0, y = y₀ + (1/x₀) $$ when y = 0, x = y₀x₀² + x₀ $$ $$ so, Area = (½).(y₀x₀² + x₀).{y₀ + (1/x₀)} = (½).(1+x₀y₀)² $$ $$ But, y₀ = 1/x₀ => x₀y₀ = 1 $$ => Area = (½).(1+1)² = 2 $$

At the point (1,1), the base=height= $\sqrt{2}$

After using the formula $A=\frac{1}{2}bh$

the final answer is 2

Since the area is given to be constant we take the simple case of x=1 so y=1 and the slope of the tangent =-1/x^2=-1 so the equatiopn of tangent is x+y=2 having intercepts of 2 units on both the axis.The area is equail to 1/2 2 2=4/2=2 sq units.