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

If $m$ is the slope of the normal to the parabola $y^2 + 4x + 4y = 0$ at the point $(-3, 2)$,which is one of the endpoints of a focal chord,find the value of $m$.

The intercepts of the focal chord (which is a part of the latus rectum) to the parabola $y^{2}=4ax$ are $b$ and $k$. Then $k$ is equal to:

The length of the latus rectum of the parabola $x^{2}+2y=8x-7$ is

Let $O$ be the vertex of the parabola $y^2 = 4x$ and its chords $OP$ and $OQ$ are perpendicular to each other. If the locus of the mid-point of the line segment $PQ$ is a conic $C$, then the length of its latus rectum is:

Find the point where the normal drawn from the upper end of the latus rectum of the parabola $y^2 = -12x$ intersects the axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module