91Ó°ÊÓ

The general equation of a straight line is given by: \(y = mx + c\) Where \(m\) is the slope and \(c\) is the y-intercept.

Now, let's find the distance of the line from the origin \((0,0)\). Using the formula for the distance between a point \((x_1, y_1)\) and a straight line \(Ax + By + C = 0\): \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\) Here, \(A = -m, B = 1, C = c\) for the line given by \(y = mx + c\). So, for the origin \((0,0)\), we have: \(d = \frac{|c|}{\sqrt{(-m)^2 + 1^2}} = \frac{|c|}{\sqrt{m^2 + 1}}\) Now, we are given that this distance is fixed, say \(p\), i.e.: \(\frac{|c|}{\sqrt{m^2 + 1}} = p\)

We can now express the y-intercept \(c\) in terms of the slope \(m\) using the above equation: \(c = \pm p\sqrt{m^2 + 1}\)

Now, substitute the value of \(c\) back into the equation of the line: \(y = mx \pm p\sqrt{m^2 + 1}\)

Next, differentiate both sides of the equation with respect to x: \(\frac{dy}{dx} = m \pm p\frac{d}{dx}(\sqrt{m^2 + 1})\) Now, we need to find \(p\frac{d}{dx}(\sqrt{m^2 + 1})\). Using the chain rule, we differentiate the term inside the square root with respect to \(m\), and then we differentiate \(m\) with respect to \(x\). Since \(\frac{dy}{dx} = m\), we first differentiate \(m\) with respect to \(x\): \(\frac{dm}{dx} = 0\) Now, differentiate \(\sqrt{m^2 + 1}\) with respect to \(m\): \(\frac{d}{dm}(\sqrt{m^2 + 1}) = \frac{2m}{2\sqrt{m^2 + 1}} = \frac{m}{\sqrt{m^2 + 1}}\) Applying the chain rule: \(p\frac{d}{dx}(\sqrt{m^2 +1}) = p\frac{dm}{dx}\frac{d}{dm}(\sqrt{m^2 + 1}) = p(0)\frac{m}{\sqrt{m^2 + 1}} = 0\) Finally, substitute this back into the differential equation: \(\frac{dy}{dx} = m\) Thus, the differential equation representing all straight lines at a fixed distance \(p\) from the origin is: \(\frac{dy}{dx} = m\)