Question

If the curve ay+x^2=7 and x^3=y, cut orthogonally at (1,1), then find the value of a

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{x^2+ay=7}\;\textsf{and}\;\mathsf{x^3=y}$

$\textsf{cut orthogonally}$

$\underline{\textsf{To find:}}$

$\textsf{The value of a}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{x^2+ay=7}$

$\textsf{Differentiate with respect to x}$

$\mathsf{2x+a\,\dfrac{dy}{dx}=0}$

$\implies\mathsf{\dfrac{dy}{dx}=\dfrac{-2x}{a}}$

$\textsf{Slope of tangent}$

$\mathsf{m_1=(\dfrac{dy}{dx})_(1,1)=\dfrac{-2}{a}}$

$\mathsf{x^3=y}$

$\textsf{Differentiate with respect to x}$

$\mathsf{3x^2=\dfrac{dy}{dx}}$

$\textsf{Slope of tangent}$

$\mathsf{m_2=(\dfrac{dy}{dx})_(1,1)=3(1)^2=3}$

$\textsf{Since the curves cut orthgonally, we have}$

$\mathsf{m_1{\times}m_2=-1}$

$\mathsf{\dfrac{-2}{a}{\times}3=-1}$

$\implies\boxed{\mathsf{a=6}}$

$\underline{\textsf{Answer:}}$

$\textsf{The value of a is 6}$