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Suppose the parabola (yk)2=4(xh), with vertex A, passes through O=(0,0) and L=(0,2). Let D be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at P. Then PDA is equal to


A
tan1119
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B
tan1219
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C
tan1419
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D
tan1819
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Solution

The correct option is A tan1219

Parabola passing through (0,0) and (0,2)
k2=4h............(i)(2k)2=4h......(ii)
Using equation (i) and (ii),
(2k)2=k244k+k2=k2k=1
Using equation(i),
h=14
Now,
tanα=m2m11+m1m2         =∣ ∣ ∣(k+2kh+1h)(k+21h+10)1+(k+2kh+1h)×(k+21h+10)∣ ∣ ∣         =∣ ∣ ∣2(k+1h+1)1+2×(k+1h+1)∣ ∣ ∣         =∣ ∣ ∣2(83)1+2×(83)∣ ∣ ∣         =219

Hence the PDA=tan1219

Mathematics

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