Find the coordinates of a point on the curve | vedclass

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

Question

(B) Given curve is $y=x^2-3x+2$. On differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x-3$. Let $m_1 = 2x-3$ be t

Vedclass · Accepted Answer

$(1,0)$

Vedclass · Answer

(B) Given curve is $y=x^2-3x+2$. On differentiating with respect to $x$,we get the slope of the tangent: $\frac{dy}{dx} = 2x-3$. Let $m_1 = 2x-3$ be the slope of the tangent at point $(x, y)$. The given line is $y=x$. Comparing this with $y=mx+c$,the slope of the line is $m_2 = 1$. Since the tangent is perpendicular to the line,the product of their slopes must be $-1$: $m_1 \cdot m_2 = -1$. $(2x-3) \cdot 1 = -1$. $2x - 3 = -1$. $2x = 2$. $x = 1$. Now,substitute $x=1$ into the curve equation to find the $y$-coordinate: $y = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$. Thus,the required point is $(1,0)$.

Find the coordinates of a point on the curve $y=x^2-3x+2$,at which the tangent drawn to this curve is perpendicular to the line $y=x$.

Similar Questions

The sum of the intercepts on the axes cut off by the tangent to the curve $x^{2/3} + y^{2/3} = 2$ at the point $(1, 1)$ is:

The equation of the tangent to the curve $y = be^{-x/a}$ at the point where it crosses the $y$-axis is:

If $y = 4x - 6$ is a tangent to the curve $y^2 = ax^4 + b$ at $(3, 6)$,then the values of $a$ and $b$ are:

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

Find the equation of all lines having slope $2$ and being tangent to the curve $y+\frac{2}{x-3}=0.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module