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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.

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    Your curve is $c$ and the equation $y=cx^2$ has same constant $c$. – 2017-01-22
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    @Ajit Kumar Gupta: your idea is right. Maybe there is some typo. – 2017-01-22
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    You are right: that question should have been formulated more carefully. – 2017-01-22
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    See the Edit I just added to my answer. – 2017-01-23
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    @JeanMarie Your answer is correct, of course, but the OP is correct too in saying that $x=0$ is also a solution. I would expect, at least, that $(0,\sqrt2)$ were not included as a possible choice. – 2017-01-23
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    I completely agree now. See the last remark I have done. – 2017-01-23

1 Answers 1

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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.

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    I got it. But why we are not getting the y-axis as another solution by using this method. For example x^2-1=0 has two solutions, we can find these solutions by using a calculative method or by seeing its graph. In both the cases (either by calculative method or by graph) we get the same (both 1 and -1) solutions, neither 1 nor -1 is missed. Similarly, we can see by graph that whenever y-axis intersects the given family, it intersects orthogonally. So it should also be another output (solution) of this method. – 2017-01-23
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    $ \dfrac{x^2}{2}+y^2=k? $ a typo maybe – 2017-01-23
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    @Narasimham Thank you very much: I spotted the equation. – 2017-01-23
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    @Ajit Kumar Gupta In fact, it is not the whole of $y$ axis, it is only point (0,0) which an ellipse reduced to a point. – 2017-01-23