The point of intersection of tangents at the | vedclass

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(B) The equation of the tangent at point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.For the given parabola $y^2 = 4x$,w

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$(-1, 0)$

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(B) The equation of the tangent at point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x + x_1)$.For the given parabola $y^2 = 4x$,we have $a = 1$.The coordinates of the ends of the latus rectum are $L(1, 2)$ and $L'(1, -2)$.The equation of the tangent at $L(1, 2)$ is $2y = 2(x + 1)$,which simplifies to $y = x + 1$.The equation of the tangent at $L'(1, -2)$ is $-2y = 2(x + 1)$,which simplifies to $y = -(x + 1)$.Solving these two equations simultaneously:$x + 1 = -(x + 1)$$2(x + 1) = 0$$x = -1$Substituting $x = -1$ into $y = x + 1$,we get $y = 0$.Thus,the point of intersection is $(-1, 0)$.

List-$I$	List-$II$
$(I)$ Vertex	$(A)$ $(-\frac{3}{2}, -3)$
$(II)$ Focus	$(B)$ $(\frac{3}{2}, -3)$
$(III)$ Equation of the directrix	$(C)$ $2x+5=0$
$(IV)$ Equation of the axis	$(D)$ $2x+y+3=0$
	$(E)$ $y+3=0$
	$(F)$ $(-2, -3)$

The point of intersection of tangents at the ends of the latus rectum of the parabola $y^2 = 4x$ is

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