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. Solution :
Plot points on the curve,
A ( x 1 , y 1 ) A(x_{1},y_{1}) A ( x 1 , y 1 ) We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.
y − y 1 = − 1 m ( x − x 1 ) y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) y − y 1 = − m 1 ( x − x 1 ) As the normal intersects the x-axis, y = 0
Substituting to the previous equation, we get
− y 1 = − 1 m ( x − x 1 ) − m y 1 = − 1 ( x − x 1 ) − m y 1 = − x + x 1 x = x 1 + m y 1 \begin{align*}
-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\
-my_{1}&=-1\left(x-x_{1}\right)\\
-my_{1}&=-x+x_{1}\\
x&=x_{1}+my_{1}
\end{align*}
− y 1 − m y 1 − m y 1 x = − m 1 ( x − x 1 ) = − 1 ( x − x 1 ) = − x + x 1 = x 1 + m y 1 By using distance formula, from the origin (0,0), to point (x1 +y1 ) = from intersection to (x1 +y1 )
( x 1 − 0 ) 2 + ( y 1 − 0 ) 2 = ( x 1 + m y 1 − x 1 ) 2 + ( 0 − y 1 ) 2 x 1 2 + y 1 2 = m 2 y 1 2 + y 1 2 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 1 y 1 ; m = d y d x x 1 = d y d x y 1 \begin{align*}
\sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\
\sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\
x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\
x_{1}^2&=m^2y_{1}^2\\
x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\
x_{1}&=\frac{dy}{dx}y_{1}
\end{align*}
( x 1 − 0 ) 2 + ( y 1 − 0 ) 2 x 1 2 + y 1 2 x 1 2 + y 1 2 x 1 2 x 1 x 1 = ( x 1 + m y 1 − x 1 ) 2 + ( 0 − y 1 ) 2 = m 2 y 1 2 + y 1 2 = m 2 y 1 2 + y 1 2 = m 2 y 1 2 = m 1 y 1 ; m = d x d y = d x d y y 1 Change x1 and y1 to x and y,
x = y d y d x x d x = y d y \begin{align*}
x&=y\frac{dy}{dx}\\
xdx&=ydy
\end{align*}
x x d x = y d x d y = y d y By integrating,
∫ y d x = ∫ x d y y 2 2 = x 2 2 + C y 2 = x 2 + 2 C \begin{align*}
\int \:ydx&=\int \:xdy\\
\frac{y^2}{2}&=\frac{x^2}{2}+C\\
y^2&={x^2}+2C\\
\end{align*}
∫ y d x 2 y 2 y 2 = ∫ x d y = 2 x 2 + C = x 2 + 2 C We get,
y 2 − x 2 − = 2 C y^2-x^2-=2C y 2 − x 2 − = 2 C
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