The point on the curve y=x^2+4x+3 which is cl | vedclass

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

Question

(B) Let the point on the curve be $P(x, y)$ where $y = x^2 + 4x + 3$. The line is $3x - y + 2 = 0$. For the point to be closest to the line,the tangen

Vedclass · Accepted Answer

$\left(-\frac{1}{2}, \frac{5}{4}\right)$

Vedclass · Answer

(B) Let the point on the curve be $P(x, y)$ where $y = x^2 + 4x + 3$. The line is $3x - y + 2 = 0$. For the point to be closest to the line,the tangent at that point must be parallel to the given line. The slope of the tangent is $\frac{dy}{dx} = 2x + 4$. Since the tangent is parallel to $y = 3x + 2$,its slope must be $3$. $2x + 4 = 3 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}$. Now,find the $y$-coordinate by substituting $x = -\frac{1}{2}$ into the curve equation: $y = \left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{5}{4}$. Thus,the point is $\left(-\frac{1}{2}, \frac{5}{4}\right)$.

The point on the curve $y=x^2+4x+3$ which is closest to the line $y=3x+2$ is

Similar Questions

The equation of the tangent to the curve $(1 + x^2)y = 2 - x$ at the point where it crosses the $x$-axis is:

Find the equations of the tangent and normal to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$ at $(1,1).$

The angle between the tangent lines to the graph of the function $f(x) = \int_{2}^{x} (2t - 5) \, dt$ at the points where the graph cuts the $x$-axis is

Find the coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=4$ at which the tangent is equally inclined to the axes.

The coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=6$ at which the tangent is equally inclined to the axes are

Vedclass Test Series

Exam Paper Generator

Online Exam Module