\[ \begin{array}{1 1}(1)45^{\circ}&(2)30^{\circ}\\(3)60^{\circ}&(4)90^{\circ}\end{array}\]

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The equation of the given parabola is $y^2=16 x$

$y^2=4(4)x=>a=4$

The equation of any tangent to the parabola $y^2=16x$ is

$y= mx +\large\frac{4}{m}$

This passes through the point $(-4,4)$

$4= -4m + \large\frac{4}{m}$

$1=-m +\large\frac{1}{m}$

$m=-m^2+1$

$m^2+m-1=0$

Let $m_1,m_2$ be the slopes of the tangents

Then product of the slopes $m_1.m_2 =\large\frac{-1}{1}$$=-1$

$90^{\circ}$ is the correct answer.

Hence 4 is the correct answer.

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