If the tangents of the parabola y^2=8x passin | vedclass

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(D) The equation of the parabola is $y^2 = 8x$,which implies $4a = 8$,so $a = 2$. Let the point $P$ be $(x_1, y_1) = (1, 3)$. The chord of contact $AB

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$\frac{1}{4}$

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(D) The equation of the parabola is $y^2 = 8x$,which implies $4a = 8$,so $a = 2$. Let the point $P$ be $(x_1, y_1) = (1, 3)$. The chord of contact $AB$ of the tangents drawn from $P(1, 3)$ to the parabola $y^2 = 8x$ is given by $yy_1 = 2a(x + x_1)$. Substituting the values,we get $3y = 2(2)(x + 1)$,which simplifies to $3y = 4x + 4$,or $4x - 3y + 4 = 0$. The area of the triangle formed by the tangents from a point $(x_1, y_1)$ to the parabola $y^2 = 4ax$ is given by the formula $	ext{Area} = \frac{(y_1^2 - 4ax_1)^{3/2}}{2a}$. Substituting $a = 2$,$x_1 = 1$,and $y_1 = 3$: $	ext{Area} = \frac{(3^2 - 4(2)(1))^{3/2}}{2(2)} = \frac{(9 - 8)^{3/2}}{4} = \frac{1^{3/2}}{4} = \frac{1}{4}$ sq. units.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

If the tangents of the parabola $y^2=8x$ passing through the point $P(1,3)$ touch the parabola at $A$ and $B$,then the area (in sq. units) of $\triangle PAB$ is

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