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Equation of a common tangent to the circle x^(2) + y^(2) - 6x = 0 and the parabola, y^(2) = 4x, is |
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Answer» `sqrt(3) y = 3x + 1` `y = mx + (a)/(m)` `:.` Equation of tangent to the parabola `y^(2) = 4x` is `y = mx + (1)/(m)``( :' a = 1)` `implies m^(2) x - my + 1 = 0` Now, let line (i) is also a tangent to the circle. Equation of circle `x^(2) + y^(2) - 6X = 0` Clearly, centre of given circle is (3, 0) and radius = 3 [`:'` for the circle `x^(2) + y^(2) + 2 gx + fy + c = 0`, centre = (-g, -f) and radius `= sqrt(g^(2) + f^(2) -c]` `:.` The perpendicular distance of (3,0) from the line (i) is 3. [`:'` Radius is perpendicular to the tangent of cirlce] `implies (|m^(2).3 - m.0 + 1|)/(sqrt((m^(2))^(2) + (-m)^(2))) = 3` The LENGTH of perpendicular from a point `(x_(1), y_(1))` to the line `ax + by + c = 0` is `|(ax_(1) + by_(1) + c)/(sqrt(a^(2) + B^(2)))|` `implies (3m^(2) + 1)/(sqrt(m^(4) + m^(2))) = 3` `implies 9m^(4) + 6m^(2) + 1 = 9 (m^(4) + m^(2))` `implies m ~~ oo` or `m = +- (1)/(sqrt(3))` `[:' underset(m to oo)("lim") (3m^(2) + 1)/(sqrt(m^(4) + m^(2))) = underset(m to oo)("lim") (3 + (1)/(m^(2)))/(sqrt(1 + (1)/(m^(2))) = 3]` `:. ` Equation of common tangents are x = 0 `y = (x)/(sqrt(3)) + sqrt(3)` and `y = (-x)/(sqrt(3)) - sqrt(3)` |
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