Question

Eqn of tangent $\cos \frac{\mathrm{t}}{2} \mathrm{y}-\mathrm{a}\left[2 \cos \frac{\mathrm{t}}{2} \sin \mathrm{t}-\sin 2 \operatorname{t} \cos \frac{\mathrm{t}}{2}\right]=$ $-\sin \frac{\mathrm{t}}{2} \mathrm{~s}+\mathrm{a}\left[2 \sin \frac{\mathrm{t}}{2} \cos \mathrm{t}+\cos 2 \mathrm{t} \sin \frac{\mathrm{t}}{2}\right]$ $\left(\cos \frac{t}{2}\right) y+\left(\sin \frac{t}{2}\right) x=a\left[\begin{array}{l}2\left[\sin \frac{t}{2} \cos t+\cos \frac{t}{2} \sin t\right] \\ +\cos 2 \operatorname{t} \sin \frac{t}{2}-\sin 2 t \operatorname{t} 0 s \frac{t}{2}\end{array}\right]$ $$ =a\left[2 \sin \frac{3 t}{2}-\sin \frac{3 t}{2}\right] $$ \(=a \sin \frac{3 t}{2} .\) so, \(p=a \sin \frac{3 t}{2}, p_{1}=3 a \cos \frac{3 t}{2}\) \(9 p^{2}+p_{1}^{2}=9 a^{2}\)

Short answer

Answer: The slope of the normal line is \(-(\cot \frac{t}{2})\).

Step-by-step solution

Convert the equation to Cartesian form

We are given an equation of the tangent line: \((\cos \frac{t}{2})y+(\sin \frac{t}{2})x=a\left[2 \sin \frac{3 t}{2}-\sin \frac{3 t}{2}\right] = a\sin \frac{3t}{2}\) Rearrange the equation to put it into the standard linear equation form: \(y = -(\tan \frac{t}{2}) x + (\frac{a}{\cos \frac{t}{2}}) \sin \frac{3t}{2}.\)

Find the slope of the tangent line and normal line

The slope of the tangent line (m_tan) is given by the coefficient of x: \(m_{tan} = -(\tan \frac{t}{2})\) Since the tangent line is perpendicular to the normal line, the product of their slopes is -1: \(m_{tan} \cdot m_{norm} = -1\) Now, find the slope of the normal line: \(m_{norm} = -(\cot \frac{t}{2})\)

Calculate the polar distance from the origin to the tangent line

The polar distance (p) from the origin to the tangent line is given by: \(p = a\sin \frac{3t}{2}\)

Find the value of p1 and verify the relation

We are given that: \(p_1 = 3a\cos \frac{3t}{2}\) Now, verify the relation \(9p^2 + p_1^2 = 9a^2\): \(9(a\sin \frac{3t}{2})^2 + (3a\cos \frac{3t}{2})^2 = 9a^2( \sin^2 \frac{3t}{2} + \cos^2 \frac{3t}{2})\) Since \(\sin^2 \theta + \cos^2 \theta = 1\) for any angle \(\theta\), the relation is verified: \(9p^2 + p_1^2 = 9a^2\)