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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 |
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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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