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