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 equation of the tangent to the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$ $(n \in N)$ at the point with abscissa equal to '$a$' is:

If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{2/3} + y^{2/3} = 4$ is parallel to the line $\sqrt{3}x + y = 1$,then $\alpha^2 + \beta^2 =$

If the normal to the curve $y = x + \frac{2}{x}$ at the point where the abscissa is $2$ meets the coordinate axes at points $A$ and $B$,find the length of $AB$.

What is the equation of the normal to the curve $y = x(2 - x)$ at the point $(2, 0)$?

If the slope of the line through $(0,0)$ which is tangent to the curve $y=x^2+x+16$ is $m$,then the value of $m-4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module