Find the condition that the curves

Solution:

It’s seen that the given curves are equation of two circles.

2x = y2 ….. (1) and

2xy = k ….. (2)

We know that, two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90o.

Now, differentiating equations (1) and (2) w.r.t. t, we get

$2.1=2 y \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{y} \Rightarrow m_{1}=\frac{1}{y}$

$\left(m_{1}=\right.$ slope of the tangent $)$

$\Rightarrow \quad 2 x y=k$

$\Rightarrow 2\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]=0$

$\therefore$ $\frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_{2}=-\frac{y}{x}$

$\left[m_{2}=\right.$ slope of the other tangent $]$

If the two tangents are perpendicular to each other,

then $\quad m_{1} \times m_{2}=-1$

$\Rightarrow \quad \frac{1}{y} \times\left(-\frac{y}{x}\right)=-1 \Rightarrow \frac{1}{x}=1 \quad \Rightarrow \quad x=1$

$2.1=2 y \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{y} \Rightarrow m_{1}=\frac{1}{y}$

$\left(m_{1}=\right.$ slope of the tangent)

$\Rightarrow$ $2 x y=k$

$\Rightarrow 2\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]=0$

$\therefore \quad \frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_{2}=-\frac{y}{x}$

$\left[m_{2}=\right.$ slope of the other tangent]

If the two tangents are perpendicular to each other,

then $\quad m_{1} \times m_{2}=-1$

$\Rightarrow \quad \frac{1}{y} \times\left(-\frac{y}{x}\right)=-1 \Rightarrow \frac{1}{x}=1 \Rightarrow x=1$

Now, solving equations (1) and (2), we have

y = k/2x [From (2)]

Putting the value of y in equation (1),

2x = (k/2x)2 ⇒ 2x = k2/4x2

8x3 = k2

8(1)3 = k2

k2 = 8

Therefore, the required condition is k2 = 8.