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Find the condition If three distinct normals can be drawn from $(h,k)$ to Parabola whose equation is given by

$$2((x-1)^2+(y-1)^2)=(x+y)^2$$

My Try:

Since its a non standard parabola i rotated the axis by $45$ degress to get rid of $xy$ term. That is

$$x=\frac{X+Y}{\sqrt{2}}$$

$$y=\frac{Y-X}{\sqrt{2}}$$

The equation got converted to as

$$X^2=\frac{4}{\sqrt{2}}\left(Y-\frac{1}{\sqrt{2}}\right)$$

Any way to proceed from here?

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I will consider a parabola $y = a x^2 + b x + c= f(x)$.

Take a point $(p,q)$ in the plane. Joint it with a point on the parabola $(x,f(x))$. This segment will be a normal to the parabola if and only if $$P_{p,q}(x)=((x,f(x))-(p,q))\cdot (1, f'(x)) = 0$$

So we are looking for $x$ satisfying the equation above, of degree $3$ in $x$. We want it to have three real solutions. The condition is that the discriminant of this cubic is $> 0$. We get a condition on $(p,q)$. The curve where this condition is $0$ is the evolute of the parabola ( see evolute).

You can see a plot of the parabola $y=x^2$ and its evolute $$27 x^2 - 16 y^3 + 24 y^2 - 12 y + 2=0$$ or, equivalently $$\frac{x^2}{2}-\left(\frac{2y-1}{3}\right)^3=0$$

From points above the evolute one can draw three normals to the parabola.

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