The equation of the tangent to the parabola y | vedclass

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(D) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.Given point $(x_1, y_1) = (a/t^2, 2

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$ty = t^2x + a$

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(D) The equation of the tangent to the parabola $y^2 = 4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x + x_1)$.Given point $(x_1, y_1) = (a/t^2, 2a/t)$.Substituting these values into the tangent equation:$y(2a/t) = 2a(x + a/t^2)$Dividing both sides by $2a$:$y/t = x + a/t^2$Multiplying both sides by $t$:$y = tx + a/t$Wait,let us re-evaluate the simplification:$y/t = (t^2x + a)/t^2$$y = (t^2x + a)/t$$ty = t^2x + a$.

The equation of the tangent to the parabola $y^2 = 4ax$ at the point $(a/t^2, 2a/t)$ is:

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