The tangent at an arbitrary point M of a curve c intersects…
The tangent at an arbitrary point $M$ of a curve $c$ intersects the $y$-axis at point $N$. Determine the curves $c$ such that the segment $MN$ is seen from the point $(a,0)$ at a right angle.
Explanation of Method
Express the $y$-intercept $N$ of the tangent, impose $\vec{AM}\cdot\vec{AN}=0$ for $A=(a,0)$, and reduce the resulting equation to linear form via $v=y^2$.
Step 1: Locate $N$ and impose perpendicularity
The tangent at $M(x,y)$ meets the $y$-axis at $N(0,y-xy')$. Perpendicularity of $\vec{AM}=(x-a,y)$ and $\vec{AN}=(-a,y-xy')$ gives $\vec{AM}\cdot\vec{AN}=0$:
\[-a(x-a)+y(y-xy')=0\quad\Longrightarrow\quad xyy'=a^2-ax+y^2.\]
Step 2: Substitute $v=y^2$
With $v=y^2$, $v'=2yy'$: dividing by $x$ and doubling, $v'-\dfrac{2v}{x}=\dfrac{2a^2}{x}-2a$.
Step 3: Solve the linear equation
With $\mu(x)=1/x^2$: $\dfrac{d}{dx}\!\left(\dfrac{v}{x^2}\right)=\dfrac{2a^2}{x^3}-\dfrac{2a}{x^2}$, so $\dfrac{v}{x^2}=-\dfrac{a^2}{x^2}+\dfrac{2a}{x}+C$, i.e. $v=Cx^2+2ax-a^2$.
Result
Final answer:
\[\boxed{\,y^2=Cx^2+2ax-a^2,\qquad C\in\mathbb{R}}\]
a family of conics (hyperbolas, parabolas, or ellipses depending on the sign of $C$).