The equation of the line passing through the point $(x^{'}, y^{'})$ and perpendicular to the line $y y^{'} = 2 a (x + x^{'})$ is

x y^{'} + 2 a y + 2 a y^{'} - x^{'} y^{'} = 0

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x y^{'} + 2 a y - 2 a y^{'} - x^{'} y^{'} = 0

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x y^{'} + 2 a y + 2 a y^{'} + x^{'} y^{'} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x y^{'} + 2 a y - 2 a y^{'} + x^{'} y^{'} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is D $x y^{'} + 2 a y - 2 a y^{'} - x^{'} y^{'} = 0$
Given line is
$y y^{'} = 2 a (x + x^{'})$
$⟹ y = \frac{2 a}{y^{'}} x + \frac{2 a x^{'}}{y^{'}}$

slope of given line $m_{1} = \frac{2 a}{y^{'}}$

Slope of required line= $\frac{- 1}{m_{1}} = \frac{- y^{'}}{2 a}$

Equation of line is given by-

$(y - y^{'}) = \frac{- y^{'}}{2 a} (x - x^{'})$

$⟹ 2 a y - 2 a y^{'} = - x y^{'} + x^{'} y^{'}$

$⟹ x y^{'} + 2 a y - 2 a y^{'} - x^{'} y^{'} = 0$

Hence the equation of required line is $x y^{'} + 2 a y - 2 a y^{'} - x^{'} y^{'} = 0$ .

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