A) 4
B) 2
C) 1
D) 3
Correct Answer: B
Solution :
The equation of any normal to y2 = 4ax at (at2, 2 at) is \[y+tx=2at+a{{t}^{3}}\] ...(i) The combined equation of the lines joining the vertex (origin) to the points of inter section of the parabola and Eq. (i) is \[{{y}^{2}}=4ax\left( \frac{y+tx}{2at+a{{t}^{3}}} \right)\] \[\Rightarrow \]\[(2t+{{t}^{3}}){{y}^{2}}=4x(y+tx)\] If Eq. (i) makes a right angle at the vertex, then coefficient of x2 + coefficient of y2 = 0. \[\therefore \]\[4t-2t-{{t}^{3}}=0\Rightarrow {{t}^{2}}=2\]You need to login to perform this action.
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