Question

If the curves $x^2 = 9A (9 - y)$ and $x^2 = A(y + 1)$ intersect orthogonally, then the value of A is

• A. 3
• B. 4
• C. 5
• D. 7

Answer 1

Answer: (B)

4

Answer 2

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