A) 30o
B) 45o
C) 60o
D) 90o
Correct Answer: D
Solution :
Here, \[P(a{{t}^{2}},\,2at)\] and S(a, 0). If the tangent at P, \[ty=x+a{{t}^{2}},\] meets the directrix \[x=-a\,\,\text{at}\,\,k,\] then \[k=\left( -a,\,\frac{a{{t}^{2}}-a}{t} \right)\] \[{{m}_{1}}=\] slope of \[SP=\frac{2at}{a({{t}^{2}}-1)}\] \[{{m}_{2}}=\] slope of \[SK=\frac{a({{t}^{2}}-1)}{-2at}\] Clearly \[{{m}_{1}}{{m}_{2}}=-1\], \[\therefore \,\angle \,PSK={{90}^{o}}.\]You need to login to perform this action.
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