The equation of the tangent to the parabola y | vedclass

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

Question

(D) Given the parabola equation $y = x^2 - x$.At $x = 1$,the $y$-coordinate is $y = (1)^2 - 1 = 0$. So,the point of tangency is $(1, 0)$.To find the s

Vedclass · Accepted Answer

$y = x - 1$

Vedclass · Answer

(D) Given the parabola equation $y = x^2 - x$.At $x = 1$,the $y$-coordinate is $y = (1)^2 - 1 = 0$. So,the point of tangency is $(1, 0)$.To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x - 1$.Evaluating the derivative at $x = 1$: $\left( \frac{dy}{dx} \right)_{x=1} = 2(1) - 1 = 1$.Using the point-slope form of the line equation $y - y_1 = m(x - x_1)$,where $m = 1$ and $(x_1, y_1) = (1, 0)$:$y - 0 = 1(x - 1)$$y = x - 1$.

The equation of the tangent to the parabola $y = x^2 - x$ at the point where $x = 1$ is:

Similar Questions

Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

If a normal chord of the parabola $y^2 = 4ax$ subtends a right angle at the vertex,find the slope of the line joining the vertex and the endpoint of the normal.

From the point $(-1, -60)$ two tangents are drawn to the parabola $y^2 = 4x$. Then the angle between the two tangents is .................. $^o$

If $\overrightarrow{AB}$ is the focal chord of the parabola $y^2=16x$ and $A=(1,-4)$,then the equation of the normal to the parabola at the point $B$ is

The length of the latus rectum of the parabola $y^2+8x-2y+17=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module