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

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(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $SS_1 = T^2$. Here,$S = y^2 - 4x$,$S_1 = (-

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$\infty$

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(C) The equation of the pair of tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by $SS_1 = T^2$. Here,$S = y^2 - 4x$,$S_1 = (-2)^2 - 4(-1) = 4 + 4 = 8$,and $T = y(-2) - 2(x - 1) = -2y - 2x + 2$. Substituting these into $SS_1 = T^2$: $(y^2 - 4x)(8) = (-2y - 2x + 2)^2$. Dividing by $4$: $2(y^2 - 4x) = (-(x + y - 1))^2 = (x + y - 1)^2$. $2y^2 - 8x = x^2 + y^2 + 1 + 2xy - 2x - 2y$. $x^2 + 2xy - y^2 + 6x - 2y + 1 = 0$. This is a pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ where $a = 1, h = 1, b = -1$. The angle $	heta$ between the lines is given by $	an 	heta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$. Since $a + b = 1 + (-1) = 0$,the denominator is $0$,which implies $	an 	heta = \infty$. Therefore,$	heta = \frac{\pi}{2}$ and $	an 	heta$ is undefined (tends to $\infty$).

List-$I$	List-$II$
$A$. Equation of the tangent drawn at $(2, \sqrt{8})$ on the curve $y^2 = 4x$ is	$(i) -36$
$B$. Equation of the normal to the curve $y^2 = 16x$,that makes an angle of $45^{\circ}$ with its axis is	$(ii) 4$
$C$. The chord joining the points $(x_1, y_1)$ and $(x_2, y_2)$ on the curve $y^2 = 12x$ is a focal chord if $y_1 y_2 =$	$(iii) 8$
$D$. $A$ value of $k$ for which $x - 3 = 0$ is the directrix of the curve $y^2 - kx + 16 = 0$ is	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

Two tangents are drawn from the point $(-1, -2)$ to the parabola $y^2 = 4x$. If $\theta$ is the angle between these tangents,then $\tan \theta = $

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