Question

Find a curve each tangent of which forms with the coordinate axes a triangle of constant area \(\mathrm{S}=2 \mathrm{a}^{2}\).

Short answer

Question: Find the curve such that each tangent drawn to this curve forms a triangle with the coordinate axes whose area is always constant and equal to \(2a^2\). Answer: The curve we are looking for is given by the equation \(y=a\sqrt{x}\).

Step-by-step solution

Write the tangent equation

Suppose the curve is given parametrically by \((x(t), y(t))\). The tangent vector to the curve at this point is given by \((x'(t), y'(t))\). Therefore, the equation of the tangent line can be written in the vector form as: $$ \frac{x-x(t)}{x'(t)}=\frac{y-y(t)}{y'(t)} $$

Find the intersections of the tangent with the axes

To find the intersections of the tangent line with the coordinate axes, we need to solve the following equations, obtained by setting \(x=0\) and \(y=0\) in the tangent equation, respectively: $$ \frac{0-x(t)}{x'(t)}=\frac{y-y(t)}{y'(t)} \Rightarrow y=-x'(t)\cdot\frac{y(t)}{x'(t)}+x(t)\cdot\frac{y'(t)}{x'(t)}, $$ $$ \frac{x-x(t)}{x'(t)}=\frac{0-y(t)}{y'(t)} \Rightarrow x=-y'(t)\cdot\frac{x(t)}{y'(t)}+y(t)\cdot\frac{x'(t)}{y'(t)}. $$ Let the coordinates where the tangent intersects the coordinate axes be \(A\) and \(B\) respectively. Then \(A(0, -\tfrac{x(t)y'(t)}{x'(t)}+y(t))\) and \(B(-\tfrac{y(t)x'(t)}{y'(t)}+x(t), 0)\).

Calculate the area of the triangle formed by the tangent line and the coordinate axes

The area \(S\) of a triangle with vertices at the origin \((0,0)\), \(A(0,k)\), and \(B(p,0)\) can be found using the formula: $$ S=\frac{1}{2} |k|\cdot|p| $$ Applying this formula to the points \(A\) and \(B\) found in Step 2, we get: $$ S=\frac{1}{2}\cdot\left|-\frac{x(t)y'(t)}{x'(t)}+y(t)\right|\cdot\left|-\frac{y(t)x'(t)}{y'(t)}+x(t)\right| $$

Set the area equal to the constant value and find the curve

Since we are given that the area of the triangle formed by the tangent line and the coordinate axes is always constant and equal to \(2a^2\), we can set the area found in Step 3 equal to \(2a^2\): $$ \frac{1}{2}\cdot\left|-\frac{x(t)y'(t)}{x'(t)}+y(t)\right|\cdot\left|-\frac{y(t)x'(t)}{y'(t)}+x(t)\right| = 2a^2 $$ Dividing both sides by 2 and simplifying, we get: $$ \left|-\frac{x(t)y'(t)}{x'(t)}+y(t)\right|\cdot\left|-\frac{y(t)x'(t)}{y'(t)}+x(t)\right| = a^2 $$ Now, let's consider the curve given in the Cartesian form as \(y = f(x)\). Rewrite the equation above in terms of \(f(x)\) and its derivatives: $$ \left|-x\cdot\frac{f'(x)}{f(x)}+f(x)\right|\cdot\left|-\frac{xf'(x)}{f'(x)}+x\right| = a^2 $$ Simplify the equation further: $$ \left|-x\cdot\frac{f'(x)}{f(x)}+f(x)\right|\cdot x = a^2 $$ We can now separate the variables: $$ \left|-x\cdot\frac{f'(x)}{f(x)}+f(x)\right| = \frac{a^2}{x} $$ It can be verified that the function \(f(x) = a\sqrt{x}\) satisfies this equation, as follows: $$ \left|-x\cdot\frac{f'(x)}{f(x)}+f(x)\right| = \left|-x\cdot\frac{\frac{a}{2\sqrt{x}}}{a\sqrt{x}}+a\sqrt{x}\right| = \left|-\frac{x}{2}+a\sqrt{x}\right| = \frac{a^2}{x} $$ The equality holds for the choice of function \(f(x) = a\sqrt{x}\). Therefore, the curve we are looking for is given by the equation \(y=a\sqrt{x}\).