The tangent and normal to the curve y = x^2 - | vedclass

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(NONE) Given the curve $y = x^2 - x + 4$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 1$.
At point $P(1, 4)$,the slope of the t

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(NONE) Given the curve $y = x^2 - x + 4$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - 1$.
At point $P(1, 4)$,the slope of the tangent is $m = \left(\frac{dy}{dx}ight)_{(1, 4)} = 2(1) - 1 = 1$.
The equation of the tangent at $(1, 4)$ is $y - 4 = 1(x - 1)$,which simplifies to $y = x + 3$.
For point $A$ on the $X$-axis,set $y = 0$: $0 = x + 3 \implies x = -3$. So,$A = (-3, 0)$.
The slope of the normal at $(1, 4)$ is $m' = -\frac{1}{m} = -1$.
The equation of the normal at $(1, 4)$ is $y - 4 = -1(x - 1)$,which simplifies to $y = -x + 5$.
For point $B$ on the $X$-axis,set $y = 0$: $0 = -x + 5 \implies x = 5$. So,$B = (5, 0)$.
The coordinates are $P(1, 4)$,$A(-3, 0)$,and $B(5, 0)$.
The base $AB$ of $\Delta PAB$ lies on the $X$-axis. Length $AB = |5 - (-3)| = 8$.
The height of $\Delta PAB$ is the $y$-coordinate of $P$,which is $4$.
Area of $\Delta PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 4 = 16$ square units.
Note: The provided solution in the input was incorrect; the correct calculation yields $16$.

The tangent and normal to the curve $y = x^2 - x + 4$ at $P(1, 4)$ intersect the $X$-axis at $A$ and $B$ respectively. Then the area of $\Delta PAB$ is .......... square units.

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