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Condition for Two Lines to Be Parallel

Two mutually ...

Question

Two mutually perpendicular tangents to the parabola $y^{2} = 4 a x$ meet the axes in $P_{1}$ and $P_{2}$ . If S is the focus of the parabola then $\frac{1}{l (S P_{1})} + \frac{1}{l (S P_{2})}$ is equal to

\frac{4}{a}

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\frac{2}{a}

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\frac{1}{a}

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\frac{1}{4 a}

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Solution

The correct option is C $\frac{1}{a}$
Let the tangents be at point

$P (a t_{1}^{2}, 2 a t_{1}), Q P (a t_{2}^{2}, 2 a t_{2})$

we know the point of intersection of tangents at any two given points

$P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2})$ is $(a t_{1} . t_{2}, 2 a (t_{1} + t_{2}))$

since the tangents are mutually perpendicular and by property of tangents Mutually perpendicular tangents to a parabola meet on the directrixi.e

$x = - a$ hence $a t_{1} . t_{2} = - a o r t_{1} . t_{2} = - 1$

Eqn of tangent to a parabola

$y = m x + \frac{a}{m}$

where m is slope of tangent

$y^{2} = 4 a x$

$2 y m = 4 a$

$m = \frac{2 a}{y}$

for $P, m_{1} = \frac{2 a}{2 a t_{1}} = \frac{1}{t_{1}}$

$y = m_{1} x + \frac{a}{m_{1}}$

$y = \frac{x}{t_{1}} + a t_{1}$

$P_{1} = p u t y = 0$

$x = - a t_{1}^{2}$

$P_{1} (- a t_{1}^{2}, 0)$

for $Q, m 2 = \frac{2 a}{2 a t_{2}} = \frac{1}{t_{2}}$

$y = \frac{x}{t_{2}} + a t_{2}$

$P_{2} = p u t y = 0$

$x = - a t_{2}^{2}$

$P 2 (- a t_{2}^{2}, 0)$

$L (S P_{1}) = a - (- a t_{1}^{2}) = a (1 + t_{1}^{2})$

$L (S P_{2}) = a - (- a t_{2}^{2}) = a (1 + t_{2}^{2})$

from 1 above

$t_{1} . t_{2} = - 1$

$\frac{1}{L (S P_{1})} + \frac{1}{L (S P_{2})} = \frac{1}{a (1 + t_{1}^{2})} + \frac{1}{a (1 + t_{2}^{2})} = \frac{1}{a} ⎛ ⎜ ⎜ ⎝ \frac{1}{1 + t_{1}^{2}} + \frac{1}{1 + \frac{1}{t_{1}^{2}}} ⎞ ⎟ ⎟ ⎠ = \frac{1}{a} (\frac{1 + t_{1}^{2}}{1 + t_{2}^{2}}) = \frac{1}{a}$

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Similar questions

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