Question

The envelope of a family of curves is a curve $f(x,y,c)=0$ whose equation is obtained by eliminating the parameter c from $f(x,y,c)=0$ and $\frac{\partial f}{\partial c}=0,$ where $\frac{\partial f}{\partial c}$ is the differential coefficient of f with respect to c, treating x and y as constants. Moreover, the envelope of the family of normals to a curve is known as the evolute of the curve. The envelope of the family of straight lines whose sum of intercepts on the axes is $4$ is:

• A. $\sqrt{x}+\sqrt{y}=2$
• B. ${(x-y)}^2-8(x+y)+16=0$
• C. ${(x-y)}^2=4(x+y)$
• D. ${(x+y)}^2=4(x-y)$

Answer 1

The correct option is A $\sqrt{x}+\sqrt{y}=2$

Let the equation of straight line be
$\frac{x}{a}+\frac{y}{b}=1$ ....(1)
Given $a+b=4$ ....(2)
Differentiating eqn (1) w.r.t. $a$ , we get
$-\frac{x}{a^2}-\frac{y}{b^2}\frac{db}{da}=0$
$\Rightarrow \frac{x}{a^2}+\frac{y}{b^2}\frac{db}{da}=0$ ....(3)
Differentiating (2) w.r.. t a, we get
$1+\frac{db}{da}=0$
$\Rightarrow \frac{db}{da}=-1$ .....(4)
From (3) and (4), we get
$\frac{x}{a^2}=\frac{y}{b^2}$ ....(5)
$\Rightarrow \frac{\frac{x}{a}}{\frac{y}{b}}=\frac{a}{b}$
Adding 1 to both sides, we get
$\Rightarrow \frac{1}{\frac{y}{b}}=\frac{4}{b}$
$\Rightarrow b^2=4y$
$\Rightarrow b=2\sqrt{y}$
Also, by eqn (5), we get
$a=2\sqrt{x}$
Now, using (2), the envelope is
$\sqrt{x}+\sqrt{y}=2$