The $y$-axis cuts the given family $y=cx^2$ orthogonally. But other option is correct (in GATE 2016). Why? Let $s$ be the curve which passes through $(0,1)$ and intersects each curve of the family $y=cx^2$ orthogonally. Then $s$ also passes through the point
$$(a)(\sqrt{2},0)\quad   (b)(0,\sqrt{2})\quad   (c) (1,1)\quad   (d) (-1,1).$$ 
The correct option is $(a)$.
But we can see that the $y$-axis cuts the given family orthogonally. And so option $(b)$ is correct. Somebody kindly tell me why I am wrong. Thanks a lot.
 A: 
Fig. 1: The orthogonal trajectories of the family of curves $y=cx^2$ is the family of homothetic ellipses $\dfrac{x^2}{2}+y^2=k.$
Step 1: Curves:
$$\tag{1}(\Gamma_c) \ \ y=c x^2$$
are solutions to a common differential equation: 
$$\tag{2} xy'=2y \ \ \iff \ \ y'=\dfrac{2y}{x}.$$
Proof: (1) can be written 
$$\tag{3}\frac{y}{x^2}=c.$$
Let us differentiate both sides of (3):
$$\dfrac{y'x-2y}{x^3}=0$$
from which (2) can be deduced.
Step 2: Recall that if a family of curves verifies a differential equation having the form $y'=\varphi(x,y)$, then the orthogonal family verifies differential equation:
$$y'=-\dfrac{1}{\varphi(x,y)}.$$
(see explanation at the bottom of this page).
In our case (see (2)), $\varphi(x,y)=\dfrac{2y}{x}$.
Thus, the differential equation of orthogonal curves to $(\Gamma_c)$ is:
$$y'=-\dfrac{x}{2y} \ \ \iff \ \ 2yy'=-x  \ \ \iff \ \ y^2=-\dfrac{x^2}{2}+k$$
i.e. a family of ellipses $(E_k)$ (see figure).
In view of the constraint that the curve must pass through point $(1,0)$, we must have $k=1$ , thus the curve is $(E_1)$ with equation:
$$\dfrac{x^2}{2}+y^2=1$$
It is thus clear that the right option (passing through point $(\sqrt{2},1)$) is option (a) (red curve).
Explanation: two curves with cartesian equations $y=f_1(x)$ and $y=f_2(x)$ are orthogonal in a certain intersection point $(x,y)$ if, at this point, their tangent vectors are orthogonal, a condition that we can write as a null dot product in the following way:
$$\pmatrix{1\\f'_1(x)} \perp \pmatrix{1\\f'_2(x)} \ \iff \ 1+f'_1(x)f'_2(x) = 0 \ \iff \ f'_2(x) = - \dfrac{1}{f'_1(x)}.$$

Edit: There is a very special ellipse which is $(E_0)$, with equation $\dfrac{x^2}{2}+y^2=0 \ \iff \ (x,y)=(0,0)$ i.e., reduced to a point.  It is what could be called an exceptional solution, but is a full solution.
