Question

The normal lines to a given curve at each point pass through (2, 0). The curve passes through (2, 3). Formulate the differential equation and hence find out the equation of the curve.

Answer 1

Hint: We should know that the equation of a normal to a curve passing through $\left( {{x_1},{y_1}} \right)$ is given by \[\left( {y - {y_,}} \right) = \dfrac{1}{{{}^ - \left( {\dfrac{{dy}}{{dx}}} \right)}}\left( {x - {x_1}} \right)\] , where $\dfrac{{dy}}{{dx}}$ is the slope of the given curve. Use this to find the required equation.

Complete step by step solution:
First we need to use the given formulae for the equation of normal to find the equation of the curve.
The equation of normal passing through point \[\left( {{x_1},{y_1}} \right)\]is
\[\left( {y - {y_1}} \right) = - \dfrac{1}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}\left( {x - {x_1}} \right)\]
Therefore, we know that the normal passes through point ( 2 , 0), so by substituting their values to the above equation we get \[\left( {y - o} \right) = - \dfrac{1}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}\left( {x - 2} \right)\]
Or we can rewrite the equation as $ - y\dfrac{{dy}}{{dx}} = \left( {x - 2} \right)$ (Differential equation)
$ - y\;dy = \left( {x - 2} \right)dx$
Now to find the equation of curve we need to integrate the above equation
\[ - \int {ydy = } \int {\left( {x - 2} \right)} \;dx\]
$ - \dfrac{y}{2} = \dfrac{{{x^2}}}{2} - 2x + c$ (after integration)
Rearrange the equation ${x^2} + {y^2} - 4x + 2c = 0$

Since 2c is also a constant it can be replaced by a constant ‘k’,
${x^2} + {y^2} - 4x + k = 0$
Since the curve passes through (2, 3) we can calculate the value of k by substituting their points to the above equation.
${\left( 2 \right)^2} + {\left( 3 \right)^2} - \left( {4 \times 2} \right) + k = 0$
$4 + 9 - 8 + k = 0$
$k = - 5$
Substituting the values of k we get ${x^2} + {y^2} - 4x - 5 = 0$
If we rearrange the equation, we get $\left( {{x^2} - 4x + 4} \right) + {y^2} - 5 - 4 = 0$
${\left( {x - 2} \right)^2} + {y^2} = 9$
This is the equation of a circle.

Note: This is direct application of differentiation and integration in the co-ordinate geometry. differentiation of y with respect to x, will not give only the rate of change of y w.r.t x. In fact, it has several meanings and we use them according to the question. For example, here we have used it as a slope. Additionally, the relation between the slopes of tangent and normal is that the product of their slopes is equal to -1.