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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) 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 tangen

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) 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.

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.

Similar Questions

What is the length of the normal at any point on the curve $y = \frac{1}{2}a(e^{x/a} + e^{-x/a})$?

If the curves $y^2 = 12x - 3$ and $y^2 = 12 - kx$ cut each other orthogonally, then the length of the sub-tangent at $(1, b)$ on the curve $y^2 = 12 - kx$ is

The tangent to a given curve $y=f(x)$ is perpendicular to the $x$-axis,if

The equation of the tangent to the curve $y = x + \frac{4}{x^2}$ which is parallel to the $X$-axis is:

The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, -\frac{4}{e}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module