The lines tangent to the curves y^3 - x^2y + | vedclass

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

Question

(D) To find the slope of the tangent at the origin $(0,0)$,we consider the lowest degree terms of the curves.For the first curve $y^3 - x^2y + 5y - 2x

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) To find the slope of the tangent at the origin $(0,0)$,we consider the lowest degree terms of the curves.For the first curve $y^3 - x^2y + 5y - 2x = 0$,the lowest degree terms are $5y - 2x = 0$.Thus,$5y = 2x$,which gives $y = \frac{2}{5}x$. The slope $m_1 = \frac{2}{5}$.For the second curve $x^4 - x^3y^2 + 5x + 2y = 0$,the lowest degree terms are $5x + 2y = 0$.Thus,$2y = -5x$,which gives $y = -\frac{5}{2}x$. The slope $m_2 = -\frac{5}{2}$.Since $m_1 \cdot m_2 = (\frac{2}{5}) \cdot (-\frac{5}{2}) = -1$,the tangents are perpendicular to each other.Therefore,the angle $	heta$ between the tangents is $\frac{\pi}{2}$.

The lines tangent to the curves $y^3 - x^2y + 5y - 2x = 0$ and $x^4 - x^3y^2 + 5x + 2y = 0$ at the origin intersect at an angle $\theta$ equal to

Similar Questions

Prove that the curves $x=y^{2}$ and $xy=k$ cut at right angles if $8k^{2}=1$.

The normal to the curve $x = a(1 + \cos \theta ), y = a \sin \theta$ at $\theta$ always passes through the fixed point

Let $C$ be a curve given by $y(x)=1+\sqrt{4x-3}$,$x>\frac{3}{4}$. If $P$ is a point on $C$ such that the tangent at $P$ has a slope of $\frac{2}{3}$,then a point through which the normal at $P$ passes is:

At which point is the tangent to the curve $y = x^3 + 5$ perpendicular to the line $x + 3y = 2$?

If the subnormal at any point on the curve $y^n = ax$ is constant,then the value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module