If the curves 2x = y^2 and 2xy = K intersect | vedclass

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

Question

(D) Given curves are $2x = y^2$ $(i)$ and $2xy = K$ $(ii)$.Solving $(i)$ and $(ii)$,substitute $2x = y^2$ into $(ii)$:$y^2 \cdot y = K \Rightarrow y^3

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given curves are $2x = y^2$ $(i)$ and $2xy = K$ $(ii)$.Solving $(i)$ and $(ii)$,substitute $2x = y^2$ into $(ii)$:$y^2 \cdot y = K \Rightarrow y^3 = K \Rightarrow y = K^{1/3}$.Then $2x = (K^{1/3})^2 = K^{2/3} \Rightarrow x = \frac{K^{2/3}}{2}$.The intersection point is $(\frac{K^{2/3}}{2}, K^{1/3})$.Differentiating $(i)$ w.r.t. $x$: $2 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{y}$.Slope $m_1$ at the intersection point is $\frac{1}{K^{1/3}}$.Differentiating $(ii)$ w.r.t. $x$: $2(y + x \frac{dy}{dx}) = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.Slope $m_2$ at the intersection point is $-\frac{K^{1/3}}{K^{2/3}/2} = -\frac{2K^{1/3}}{K^{2/3}} = -2K^{-1/3}$.Since the curves intersect perpendicularly,$m_1 \cdot m_2 = -1$.$(\frac{1}{K^{1/3}}) \cdot (-2K^{-1/3}) = -1$.$-2K^{-2/3} = -1 \Rightarrow K^{-2/3} = \frac{1}{2}$.$K^{2/3} = 2 \Rightarrow (K^{2/3})^3 = 2^3 \Rightarrow K^2 = 8$.

If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly,then the value of $K^2$ is

Similar Questions

The abscissa of the point on the curve $y=a\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)$ where the tangent is parallel to the $X$-axis is

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$ is

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

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

At what points of the curve $y = \frac{2}{3}x^3 + \frac{1}{2}x^2$ does the tangent make an equal angle with the axes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module