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)

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 :
$x^2=9A(9-y)$ ....(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 $2x=A\frac{dy}{dx}\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{2x_1}{A}$
$\Rightarrow m_2=\frac{2x_1}{A}$
$m_1{m}_2=-1\Rightarrow \frac{4x^2}{9A^2}=1\Rightarrow 4x_1^2=9A^2$ ....(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