If the curve ay + x² = 7 and x³ = y, cut orthogonally at (1, 1) then the value of a is

The correct option is 6.

To find the value of $a$ such that the curves cut orthogonally at the point $(1,1)$, we need to understand the condition for orthogonality.

Two curves intersect orthogonally if the tangents to the curves at their point of intersection are perpendicular to each other. This means that the product of the slopes of the tangents at the intersection point is $-1$.

Let $m_1$ be the slope of the tangent to the first curve, and $m_2$ be the slope of the tangent to the second curve at $(1,1)$. The condition for orthogonality is:
$m_1\cdot m_2=-1$

First, let's find the slope $m_1$ of the first curve: $ay+x^2=7$.
Differentiating both sides with respect to $x$:
$a\frac{dy}{dx}+2x=0$
$a\frac{dy}{dx}=-2x$
$\frac{dy}{dx}=\frac{-2x}{a}$

Now, evaluating this slope at the intersection point $(1,1)$:
$m_1={\left[\frac{dy}{dx}\right]}_{(1,1)}=-\frac{2}{a}$

Next, let's find the slope $m_2$ of the second curve: $x^3=y$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx}=3x^2$

Evaluating this slope at the intersection point $(1,1)$:
$m_2={\left[\frac{dy}{dx}\right]}_{(1,1)}=3{\left(1\right)}^2=3$

Since the curves cut orthogonally, we apply the condition $m_1\cdot m_2=-1$:
$\left(-\frac{2}{a}\right)\cdot 3=-1$
$-\frac{6}{a}=-1$
$\frac{6}{a}=1$
$a=6$

Thus, the value of $a$ is $6$.