. We have, dy . Hence at ( , ), dx = Therefore, the required equation of tangent is ⇒ . The slope of the normal at the point ( , ) is − .
Therefore, the required equation of normal is = − ⇒ . Example . Find the points on the curve y at which the tangent is parallel to the line y The slope of the line y is . The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also .
Hence, dx = x + = which gives x = . Hence, x = and x = . Therefore, at ( , – ) and ( , – ) the tangent is parallel to the line y Example . Find the equation of the tangent and normal at any point to the Lissajous curve given by = and y t t .
Observe that the given curve is neither a circle nor an ellipse. For your reference the curve is shown in Fig. . .
Now, dy dx = dt dxdt = − = − t . Therefore, the tangent at any point is − = − t x That is, = Fig. . Lissajous curve t y ; Fig.
. ( , ) −− - - Applications of Differential Calculus The slope of the normal is the negative of the reciprocal of the tangent which in this case is t . Hence, the equation of the normal is t x ) . That is, x