If the tangent to the curve y^2 = 4x at point | vedclass

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(A) The equation of the parabola is $y^2 = 4x$.Comparing with $y^2 = 4ax$,we get $a = 1$.The equation of the tangent at point $(x_1, y_1) = (1, 2)$ is

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$0.5$

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(A) The equation of the parabola is $y^2 = 4x$.Comparing with $y^2 = 4ax$,we get $a = 1$.The equation of the tangent at point $(x_1, y_1) = (1, 2)$ is given by $yy_1 = 2a(x + x_1)$.Substituting the values,we get $2y = 2(1)(x + 1)$,which simplifies to $y = x + 1$.To find the points $A$ and $B$ where the tangent cuts the coordinate axes:For the $x$-axis,set $y = 0$: $0 = x + 1 \implies x = -1$. So,point $A$ is $(-1, 0)$.For the $y$-axis,set $x = 0$: $y = 0 + 1 \implies y = 1$. So,point $B$ is $(0, 1)$.The triangle $\Delta AOB$ is a right-angled triangle with vertices at $O(0, 0)$,$A(-1, 0)$,and $B(0, 1)$.The base $OA = |-1| = 1$ and the height $OB = |1| = 1$.Area of $\Delta AOB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = 0.5$.

If the tangent to the curve $y^2 = 4x$ at point $(1, 2)$ cuts the coordinate axes at points $A$ and $B$,then the area of $\Delta AOB$ is (where $'O'$ is the origin).

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