The curve passing through the point $(1, 2)$ that cuts each member of the family of parabolas $y^{2} = 4 a x$ orthogonally is :

2 x^{2} + y^{2} = 6

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x^{2} + y^{2} = 5

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x^{2} + 2 y^{2} = 9

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y^{2} - x^{2} = 3

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Solution

The correct option is A $2 x^{2} + y^{2} = 6$
$y^{2} = 4 a x . . (1)$
$\Rightarrow \frac{d y}{d x} = \frac{2 a}{y} = \frac{2 a y}{y^{2}} = \frac{2 a y}{4 a x} = \frac{y}{2 x} = m$
Now equation of orthogonal curve of $(1)$ is
$\frac{d y}{d x} = - 1 / m = \frac{- 2 x}{y}$
$\Rightarrow y d y + 2 x d x = 0$
Integrating we get
$\Rightarrow y^{2} + 2 x^{2} = c^{2}$ where $c^{2}$ is any constant
Now given this curve is passing through $(1, 2)$
$\Rightarrow 2^{2} + 2 (1)^{2} = c^{2} \Rightarrow c^{2} = 6$
Hence equation of required curve is
$2 x^{2} + y^{2} = 6$

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The curve passing through the point(1,2) that cuts each member of the family of parabolas y2=4ax orthogonally is :

The curve passing through the point $(1, 2)$ that cuts each member of the family of parabolas $y^{2} = 4 a x$ orthogonally is :