Let $\gamma$ be the curve which passes through $(0,1)$ and intersects each curve of the family $y=cx^2$ orthogonally. Then show that $γ$ also passes through the point $(\sqrt 2,0)$.
I am unable to understand is the question demanding to find the equation of the tangent to the curve $y=x^2$ because here the curve meets each curve of the family orthogonally.
I am unable to understand what to do in the given problem.Please give some hints
Suppose $\gamma$ passes through the point $(x,y)=(x,cx^2)$. At that point, the derivative of $\gamma$ has to be orthogonal to the derivative of $y=cx^2$, the latter of which is $2cx$: $$\frac{dy}{dx}=-\frac1{2cx}=-\frac x{2cx^2}=\frac{-x}{2y}$$ We've got a separable differential equation. Solving: $$2y\ dy=-x\ dx$$ $$\int2y\ dy=\int-x\ dx$$ $$y^2=-\frac12x^2+K$$ Since $\gamma$ passes through $(0,1)$: $$1^2=-\frac120^2+K\implies K=1$$ Hence the equation of $\gamma$ is $$y^2=-\frac12x^2+1$$ Since $(x,y)=(\sqrt2,0)$ satisfies this equation, $\gamma$ passes through $(\sqrt2,0)$ as well.