# Finding the curve that matches specific requirements.

Find the curve whose distance of every tangent from origin is equal to the abscissa of point of intersection of curve and tangent. Find this curve by appointing and solving differential equation needed.

I managed to draw this image and i am supposed to find $y$ such that blue line is equal to the yellow one. Now, i should derive some equations from this picture in order to determine the equation of the curve. Only thing i found up to now is equation of tangent $(y-y_0)=y'(x_0)(x-x_0)$ but this is nowhere near the solution, am i supposed to see more from this picture?

• This question is highly unusual. The blue line and yellow line can only be the same length if the blue line is the yellow line (because the yellow line is otherwise always a leg of a right triangle for which the blue line is the hypotenuse), which means all points of tangency must be on the $x$-axis, which means all (differentiable) points of the function are on $y = 0$, which means the function is $f(x) = 0$. I wonder if "distance of every tangent from origin" is supposed to be interpreted differently? Perhaps it refers to the shortest distance between the tangent line and the origin? – tilper Jun 7 '17 at 17:46
• @tilper I agree, usually you'd interpret that as the distance from a point to the line, so the blue segment should be orthogonal to the red line. In that case, from the equation of the tangent you can easily express the distance from the origin to the tangent, and equate it to $x_0$. – N.Bach Jun 7 '17 at 17:56
• Sorry guys, i draw incorrect picture, more precisely, i misinterpreted distance of tangent from origin, it is supposed to be distance orthogonal to the origin, as @Chappers answered below – cdummie Jun 8 '17 at 12:05

So, consider the tangent at $x$. The equation of the tangent is $$Y = y(x) + y'(x)(X-x).$$ We want the closest point on this to the origin. The best way to do this is probably exploit similar triangles: if $O=(0,0)$, $A=(x,0)$, $B=(x,y)$, $C=(0,y-xy')$ is the $y$-intercept of the tangent, $D$ the closest point on the tangent to the origin, and $E=(x-y/y',0)$ the $x$-intercept of the tangent, then $$\triangle ABE$$ is similar to $\triangle DCO$, so $$\frac{EA}{EB} = \frac{CO}{DO} = \cos{\theta} = \frac{1}{\sqrt{1+y'^2}}$$ (where $\theta=\angle BEA$ so $\tan{\theta}=y'$), and so we want $$DO = x \implies x = \frac{y-xy'}{\sqrt{1+y'^2}}$$ or, assuming $y>0$, $$x^2(1+y'^2) = (y-xy')^2 = y^2-2xyy'+x^2y'^2 \\ x^2-y^2+2xyy' = 0,$$ so we get $$y'=\frac{y^2-x^2}{2xy},$$ which has solutions $$y^2 + x^2 = Ax,$$ i.e. which are circles tangent to the $y$-axis.
• Mm, good point. Luckily the rest of the diagram is symmetric on reversing the rôles of $y$ and $x$, so the answer is still a circle. Edited. – Chappers Jun 8 '17 at 12:28