If the normal to a parabola y2 - 4ax at P meets the curve again at Q and if PQ and the normal at Q makes angle - and - respectively with the x-axis then tan-tan-tan- has the value equal to

Y^2 =4ax Point on the parabola (at, 2at) Slope of the tangent, 2y y' = 4a ⇒ y' = 1t Slope of the normal, m_n = t So, tan α= t_1 and tan β= t_2 Also, t_2= t_1 ...

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Standard XIII

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If the normal...

Question

If the normal to a parabola $y^{2} = 4 a x$ at $P$ meets the curve again at $Q$ and if $P Q$ and the normal at $Q$ makes angle $α$ and $β$ , respectively with the $x$ -axis then $tan α (tan α + tan β)$ has the value equal to

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Solution

The correct option is B $- 2$
$y^{2} = 4 a x$

Point on the parabola $(a t, 2 a t)$
Slope of the tangent,
$2 y y^{'} = 4 a \Rightarrow y^{'} = \frac{1}{t}$
Slope of the normal,
$m_{n} = - t$
So,
$tan α = - t_{1}$ and $tan β = - t_{2}$
Also, $t_{2} = - t_{1} - \frac{2}{t_{1}}$
$t_{1} t_{2} + t_{1}^{2} = - 2$
$tan α tan β + {tan}^{2} α = - 2$

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If the normal to a parabola y2=4ax at P meets the curve again at Q and if PQ and the normal at Q makes angle α and β, respectively with the x-axis then tanα(tanα+tanβ) has the value equal to

The correct option is B −2y2=4ax Point on the parabola (at,2at) Slope of the tangent, 2yy′=4a⇒y′=1t Slope of the normal, mn=−t So, tan α=−t1 and tan β=−t2 Also, t2=−t1−2t1 t1t2+t21=−2 tanα tanβ+tan2α=−2

If the normal to a parabola $y^{2} = 4 a x$ at $P$ meets the curve again at $Q$ and if $P Q$ and the normal at $Q$ makes angle $α$ and $β$ , respectively with the $x$ -axis then $tan α (tan α + tan β)$ has the value equal to