The two curves x=y^2 and xy=a^3 cut orthogona | vedclass

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

Question

(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).For curve $(i)$, differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).For curve $(i)$, differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.For curve (ii), differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.To find the point of intersection, substitute $x=y^2$ into $xy=a^3$: $y^2(y) = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.Then $x = a^2$. So the point of intersection is $(a^2, a)$.The slopes at $(a^2, a)$ are $m_1 = \frac{1}{2a}$ and $m_2 = -\frac{a}{a^2} = -\frac{1}{a}$.Since the curves cut orthogonally, $m_1 m_2 = -1$.Therefore, $(\frac{1}{2a})(-\frac{1}{a}) = -1 \Rightarrow -\frac{1}{2a^2} = -1 \Rightarrow 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

The two curves $x=y^2$ and $xy=a^3$ cut orthogonally at a point, then $a^2$ is equal to

Similar Questions

What is the correct order of occurrence ($\%$ by weight) in air of $Ne, Ar$ and $Kr$?

$A$ coordination complex contains $Co^{3+}$,$Cl^{-}$,and $NH_3$. When dissolved in water,one mole of this complex gives a total of $3$ moles of ions. The complex is

Which artery is absent in a frog?

Which of the following statements is false regarding the properties of electromagnetic waves?

$A$ magnetized wire of magnetic moment $M$ and length $L$ is bent into an arc of a circle subtending an angle of $60^{\circ}$ at the center. The new magnetic moment is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module