1.

Locus of the point of intersection of the tangents to the parabola y2 = 4ax which include 60° angle is(A)   y2 = 4ax + 3(x + a)2(B)   y2 - 4ax = 2(x + a)2(C)   y2 - 4ax = (x + a)2(D)   y2 - 4ax = 3(x + a)3

Answer»

Correct option (A)  y2 = 4ax + 3(x+ a)2

Explanation :

According to Problem 28 of the section ‘Subjective Problems’, the locus of the point of intersection of the tangents to the parabola y2 = 4ax which include angle α is

cot2 α(y2 - 4ax) = (x +  a)2

Now, substituting  α = 60° the required locus is

1/3(y - 4ax) = (x + a)2

y2 - 4ax = 3(x + a)2



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