Find the curve of a family of curves with the following conditions

Question

Let $\mathfrak{J}$ be the family of curves such that their normal line at each point is tangent to the parabola of equation $$y=kx^2$$ that passes through that point. Find the curve $C\in\mathfrak{J}$ that passes through $(0,1)$.

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My work: $$\mathfrak{J}\colon y=kx^2\to k=\frac{y}{x^2},\quad y'=2kx=2\frac{y}{x^2}x=2\frac{y}{x}.$$ Moreover, $$\mathfrak{J}^\perp\colon -\frac{1}{y'}=2\frac{y}{x}\to-\frac x2=yy'\to\frac{y^2}2=-x^2+c\to y^2=-2x^2+C.$$ Since $y(0)=1$, we have that $1=C$, so the final answer is $$\boxed{y^2=-2x^2+1}$$

Is it correct?

The answer does not match with the book, the answer should be:

$x^2+2y^2=2$.

Answer 1

This line in your attempt is correct: $$yy'=-\frac x 2$$ Then you have : $$ 2yy'=-x \implies (y^2)'=-x$$ Integrate: $$y^2=-\frac {x^2}2+c$$ $$2y^2+ {x^2}=2c$$ $$2y^2+ {x^2}=2$$