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Q.
If the curves $x^2 = 9A (9 - y)$ and $x^2 = A(y + 1)$ intersect orthogonally, then the value of A is
Application of Derivatives
Solution:
If two curves intersect each other orthogonally, then the slopes of corresponding tangents at the point of
intersection are perpendicular.
Let the point of intersection be $(x_1, y_1)$
Given curves :
$x^2 = 9 \, A \, (9 - y)$ ....(1)
and $x^2 = A \, (y + 1)$ ....(2)
Differentiating w.r. to x both sides equations (1) and (2) respectively, we get
$ 2x =- 9A \frac{dy}{dx} $
$ \Rightarrow \left(\frac{dy}{dx}\right)_{\left(x_1, y_1\right)} = - \frac{2x_{1}}{9A} \Rightarrow m_{1} = - \frac{2x_{1}}{9A} $
and $2x = A \frac{dy}{dx} \Rightarrow \left(\frac{dy}{dx}\right)_{\left(x_1, y_1\right)} = \frac{2x_{1}}{A} $
$\Rightarrow m_{2} = \frac{2x_{1}}{A} $
$ m_{1}m_{2} = - 1 \Rightarrow \frac{4x^{2}}{9A^{2}}= 1 \Rightarrow 4x_{1}^{2} = 9 A^{2} $ ....(3)
Solving equations (1) and (2),
we find $y_1 = 8$
Substituting $y_1 = 8$ in equation (2),
we get $x_1^2 = 9A $ ....(4)
From equations (3) and (4), we get A = 4