If the tangents to the parabola $y_{2} = 4 a x$ at the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ meet at the point $(x_{3}, y_{3})$ , then

y_{3} = \sqrt{y_{1} y_{2}}

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2 y_{3} = y_{1} + y_{2}

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\frac{2}{y_{3}} = \frac{1}{y_{1}} + \frac{1}{y_{2}}

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None of these

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Solution

The correct option is B $2 y_{3} = y_{1} + y_{2}$
$\frac{y_{2} - y_{3}}{x_{2} - x_{3}} = \frac{2 a}{y_{2}}$
$\frac{y_{1} - y_{3}}{x_{1} - x_{3}} = \frac{4 a}{2 y_{1}}$
$(y_{1} - y_{3}) y_{1} = 2 a (x_{1} - x_{3})$
$(y_{2} - y_{3}) y_{2} = 2 a (x_{2} - x_{3})$
$x_{1} - \frac{(y_{1} - y_{3}) y_{1}}{2 a} = x_{2} \frac{(y_{2} - y_{3}) y_{2}}{2 a}$
$\frac{y^{2}}{4 a} - \frac{(y_{1} - y_{3}) y_{1}}{2 a} = \frac{y_{2}^{2}}{4 a} - \frac{(y_{2} - y_{3}) y_{2}}{2 a}$
$y_{1}^{2} - 2 y_{1}^{2} + 2 y_{3} y_{1} = y_{2}^{2} - 2 y_{2}^{2} + 2 y_{3} y_{2}$
$- y_{1}^{2} + 2 y_{3} y_{1} = - y_{2}^{2} + 2 y_{3} y_{2}$
$y_{1} (2 y_{3} - y_{1}) = y_{2} (2 y_{3} - y_{2})$
$2 y_{3} (y_{1} - y_{2}) = (y_{1} + y_{2}) (y_{1} - y_{2})$
$2 y_{3} = y_{1} + y_{2}$

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