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

$\begin{array}{1 1} 1 \\ 0 \\ -6 \\ 6 \end{array} $

1 Answer

  • Slope of a line is $ ax+by+c=0$ is $ - \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
  • If $ y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p = $ slope of the normal to $ y = f(x)$ at point $p$
  • If the tangent of the curves touch orthogonally then the product of their slopes is -1.
Step 1
Equation of the given curves are
$ ay+x^2=7\: and \: x^3=y$
Consider the equation,
$ ay+x^2=7$
differentiating w.r.t $x$ we get,
$ a.\large\frac{dy}{dx}+2x=0$
$ \Rightarrow \large\frac{dy}{dx}=-\large\frac{2x}{a}$
Let this be $ m_1 \Rightarrow m_1=-\large\frac{2x}{a}$
$ \Rightarrow m_1\: at \: (1,1)=-\large\frac{2}{a}$
Consider the equation,
differentiating w.r.t $x$
$ \large\frac{dy}{dx}=3x^2$
Let this be $ m_2 \Rightarrow m_2 \: at \: (1,1)=3$
Step 2
Since the tangents touch orthogonally,
$ m_1 \times m_2 = -1$
$ -\large\frac{2}{a} \times 3 = -1$
$ \Rightarrow a = 6$
The correct option is D
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