Answer to Question #100426 in Differential Equations for Fatimah Hasana

Question

Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.

Answer 1

We need to Find the equation of the curve.

Let a Point on the curve as $A(x_1, y_1)$

Let Slope of the Tangent at this point = m

Slope of normal at this point is $= \frac {-1}{m}$

So, Equation of the normal is

$$y - y_1 = \frac {-1}{m} (x - x_1)$$

Normal intersecting the x axis, so $y = 0$

So, the above equation becomes

$$- y_1 = \frac {-1}{m} (x - x_1)$$

$$- m y_1 = -1(x - x_1) \\ x = x_1 + my_1$$

So, the point is $(x_1 + my_1, 0)$

Origin =(0,0)

Taken Point = $(x_1, y_1)$

Distance from origin to $(x_1, y_1)$ = Distance from Intersection point with normal to $(x_1, y_1)$

$$\sqrt {( x_1 - 0)^2 +(y_1 - 0)^2} = \sqrt {(x_1+my_1-x_1)^2+ (0- y_1)^2}$$

$$\sqrt {x_1^2 +y_1^2} = \sqrt {m^2 y_1^2 + y_1^2}$$

Squaring on both sides

$${x_1^2 +y_1^2} = {m^2 y_1^2 + y_1^2}$$

$${x_1^2 } = {m^2 y_1^2 }$$

$$x_1 = m y_1$$

Here,

$$m = \frac {dy}{dx}$$

$$x_1 = \frac {dy}{dx} y_1$$

Transform $(x_1, y_1) \space to \space (x,y)$

$$x = y \frac {dy}{dx}$$

$$y \space dy = x \space dx$$

Integrating,

$$\int y dy = \int x dx + c$$

Here, c is constant

$$\frac {y^2}{2} = \frac {x^2}{2} + c$$

$$\frac {y^2}{2} - \frac {x^2}{2} =c$$

$$y^2 - x^2 = 2c$$

Answer:

the equation of the curve is $y^2 - x^2 = 2c$ .