Tangents And Normals

Concept: Calculus - Derivatives of Parametric Functions and Application of Derivatives (Tangents and Normals).

Step 1: Find the derivatives of and with respect to . Given , apply the product rule to differentiate: . Given , differentiate similarly: .

Step 2: Determine the slope of the normal to the curve. First, find the slope of the tangent, which is . Using the parametric derivative formula, . The slope of the normal ( ) is the negative reciprocal of the tangent's slope: .

Step 3: Write the equation of the normal line. The point slope form of a line is . Let be general coordinates on the line. Substitute the coordinates and the slope of the normal: .

Step 4: Simplify the equation of the normal into standard form. Cross-multiply by : . The terms cancel out on both sides. Move to the left side and to the right side: . Factor out the 2 to get . Since , the standard equation of the normal line is .

Step 5: Calculate the perpendicular distance from the origin. The perpendicular distance from the origin to the line is given by .
Substitute our line's coefficients ( , , ): units. $ $

Answer (c) (2,1)

Concept: The equation of the tangent to a curve at is given by: where:

• is the slope of the tangent obtained using differentiation.
• is the point on the curve where the tangent touches.

Thus, we first find the derivative to determine the slope and then use the point–slope form of a straight line.
Step 1: {Differentiate the given function.} Given Differentiating with respect to : This represents the slope of the tangent at any point .
Step 2: {Find the slope at .} Thus, the slope of the tangent is .
Step 3: {Find the point on the curve at .} Substitute into the original equation: So the point of tangency is
Step 4: {Use the point–slope form of the tangent line.} Substituting , , :