An equation of the tangent to the curve y = x | vedclass

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

Question

(A) Let the point of contact be $(h, k)$,where $k = h^4$. The slope of the tangent to the curve $y = x^4$ at $(h, k)$ is given by $\frac{dy}{dx} = 4x^

Vedclass · Accepted Answer

$y = 0$

Vedclass · Answer

(A) Let the point of contact be $(h, k)$,where $k = h^4$. The slope of the tangent to the curve $y = x^4$ at $(h, k)$ is given by $\frac{dy}{dx} = 4x^3$. At $x = h$,the slope is $m = 4h^3$. The equation of the tangent at $(h, k)$ is $y - k = 4h^3(x - h)$. Since the tangent passes through the point $(2, 0)$,we substitute these coordinates into the equation: $0 - h^4 = 4h^3(2 - h)$ $-h^4 = 8h^3 - 4h^4$ $3h^4 - 8h^3 = 0$ $h^3(3h - 8) = 0$ This gives $h = 0$ or $h = \frac{8}{3}$. For $h = 0$,the equation of the tangent is $y - 0 = 4(0)^3(x - 0)$,which simplifies to $y = 0$. For $h = \frac{8}{3}$,the equation of the tangent is $y - (\frac{8}{3})^4 = 4(\frac{8}{3})^3(x - \frac{8}{3})$. Thus,$y = 0$ is one of the equations of the tangent.

An equation of the tangent to the curve $y = x^4$ from the point $(2, 0)$ not on the curve is

Similar Questions

At which point on the curve $y^2 = x$ does the tangent make an angle of $45^{\circ}$ with the $x-$axis?

Show that the tangents to the curve $y=7x^3+11$ at the points where $x=2$ and $x=-2$ are parallel.

Find a point on the curve $y=(x-2)^{2}$ at which the tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$.

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to

If the tangent at point $(1, 2)$ on the curve $y = ax^2 + bx + \frac{7}{2}$ is parallel to the normal at $(-2, 2)$ on the curve $y = x^2 + 6x + 10$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module