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Question

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

A
0
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B
2
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C
12
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D
1
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Solution

The correct option is B 2
Let the given equation of parabola is
y2=4ax

Let the normal at P(at12,2at1) meets the parabola again at Q(at22,2at2).

Equation of normal at P is given by,
y=t1x+2at1+at31

Equation of normal at Q is given by,
y=t2x+2at2+at32

So, tanα=t1 and tanβ=t2

Also, t2=t12t1

t1t2+t21=2

tanαtanβ+tan2α=2

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