The area of the triangle formed by the tangen | vedclass

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(D) Given the curve $y^2=4x$ and the point $P(1,2)$.The slope of the tangent at $(1,2)$ is found by differentiating $y^2=4x$: $2y \frac{dy}{dx} = 4 \R

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(D) Given the curve $y^2=4x$ and the point $P(1,2)$.The slope of the tangent at $(1,2)$ is found by differentiating $y^2=4x$: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$. At $(1,2)$,the slope $m_t = \frac{2}{2} = 1$.The equation of the tangent is $y-2 = 1(x-1) \Rightarrow y = x+1$.The tangent intersects the $Y$-axis $(x=0)$ at $A(0,1)$.The slope of the normal $m_n = -\frac{1}{m_t} = -1$.The equation of the normal is $y-2 = -1(x-1) \Rightarrow y = -x+3$.The normal intersects the $Y$-axis $(x=0)$ at $B(0,3)$.The triangle is formed by vertices $P(1,2)$,$A(0,1)$,and $B(0,3)$.The base of the triangle along the $Y$-axis is the distance between $A(0,1)$ and $B(0,3)$,which is $|3-1| = 2$ units.The height of the triangle is the perpendicular distance from $P(1,2)$ to the $Y$-axis,which is the $x$-coordinate of $P$,i.e.,$1$ unit.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 1 = 1$ square unit.

The area of the triangle formed by the tangent and the normal drawn to the curve $y^2=4x$ at $(1,2)$ with the $Y$-axis is (in square units):

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