How to find locus for given condition?

Question

I've been through all the thinking but i could not get to the answer. Please help me in this. See images for details

Answer 1

There is no other parabola other than $C_2$ having the given condition. The argument goes as follows $:$

If there exists such a $C_1$ then since both $C_1$ and $C_2$ have the same latus rectum the general equation of $C_1$ is given by $$y=ax^2+bx+c\ \ \ \ \ \ (1)$$ where $a \neq 0.$ Since $C_1$ touches $C_2$ at the origin so we can say that $C_1$ passes through the origin. Therefore the general equation of $C_1$ reduces to $$y=ax^2+bx\ \ \ \ \ \ (2)$$ where $a \neq 0.$ Since $C_1$ and $C_2$ have the same latus rectum so we can also say that $C_1$ passes through the end points of the latus rectum of $C_2$ which are $\left (- \frac 1 2,-\frac 1 4 \right )$ and $\left ( \frac 1 2,-\frac 1 4 \right ).$ So they will satisfy $(2).$ Therefore we get two equations in two unknowns $$\begin{align*} a-2b & =-1 \\ a+2b& = -1 \end{align*}$$ Solving we get $b=0$ and $a=-1.$ So $C_1$ is of the form $$x^2 = -y.$$ This shows that $C_1$ coincides with $C_2.$

So the locus of focus of $C_1$ is simply the focus of $C_2,$ which is $\left (0,-\frac 1 4\right ).$ So the given locus $x^2+y^2=2ay^4$ should be satisfied by this point. Solving you will end up with $a=8.$

But you can see that if we put $a=8$ then the locus takes the form $$x^2+y^2 -16y^4 = 0$$ which doesn't represent the point $\left (0,-\frac 1 4 \right )$ only because it is also satisfied by the point $(0,0).$

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Conclusion $:$ The question has been wrongly posed.