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The ends of a straight line segment of constant length AB = a slide on the axes of rectangular coordinate system. The straight line AC and BC are parallel to the coordinate axes and they intersect each other at point C. Find the equation of the geometrical locus of the foot M of the perpendicular drawn from C on the straight line AB.

Please help!

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Here's the derivation in the first quadrant. With points labeled as in the (slightly tilted) diagram, parameterize the point $M=(x(\theta),y(\theta))$ with the angle $\theta:=\angle OBA=\angle MCB$. Using trig we establish $$ h=a\sin\theta,\quad w=a\cos\theta $$ and $$MB=h\sin\theta=a\sin^2\theta.$$ Drop a perpendicular from $M$ to meet $OB$ at point $N$. Then $$ y = MN = MB \sin \theta = a\sin^3\theta $$ and $$w - x = NB = MB\cos\theta=a\sin^2\theta\cos\theta$$ which gives the parameterization for $(x,y)$: $$x=a\cos^3\theta,\quad y=a\sin^3\theta.$$

If you eliminate the parameter $\theta$ you'll get the equation $$\left(\frac xa\right)^{2/3} + \left(\frac ya\right)^{2/3} = 1.$$

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Let $OA=m$ and $OB=n$ then $m^2+n^2=a^2$.

Equation of $AB$: $\dfrac{x}{m}+\dfrac{y}{n}=1$ or $nx+my=mn$

Slope of $AB$: $-\dfrac{n}{m}$

Slop of $CM$: $\dfrac{m}{n}$

Coordinates of $C$: $(m,n)$

Equation of $MC$: $y=\dfrac{m}{n}(x-m)+n$ or $ny-mx=-m^2+n^2$.

Contact $AB$ and $MC$ in $M$: \begin{cases} nx+my=mn,\\ ny-mx=-m^2+n^2. \end{cases} so $x=\dfrac{m^3}{m^2+n^2}$ and $y=\dfrac{n^3}{m^2+n^2}$ these give us $m=\sqrt[3]{a^2x}$ and $n=\sqrt[3]{a^2y}$ with $m^2+n^2=a^2$ we conclude $$\color{blue}{x^\frac23+y^\frac23=a^\frac23}$$.

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