Two tangents are drawn from a point (-2, -1) | vedclass

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(D) The locus of the point of intersection of tangents to the parabola $y^2 = 4ax$ inclined at an angle $\alpha$ to each other is given by the equatio

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$3$

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(D) The locus of the point of intersection of tangents to the parabola $y^2 = 4ax$ inclined at an angle $\alpha$ to each other is given by the equation: $	an^2 \alpha (x + a)^2 = y^2 - 4ax$ Given the equation of the parabola $y^2 = 4x$,we have $a = 1$. The point of intersection is $(x, y) = (-2, -1)$. Substituting these values into the locus equation: $	an^2 \alpha (-2 + 1)^2 = (-1)^2 - 4(1)(-2)$ $	an^2 \alpha (-1)^2 = 1 + 8$ $	an^2 \alpha (1) = 9$ $	an^2 \alpha = 9$ Taking the square root on both sides: $|	an \alpha| = 3$

Two tangents are drawn from a point $(-2, -1)$ to the curve $y^2 = 4x$. If $\alpha$ is the angle between them,then $|\tan \alpha|$ is equal to:

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