107. Show that the normal at any point θ to the curve x = a cosθ + a θ sin θ, y = a sinθ – aθ cosθ is at a constant distance from the origin.

We have $x=acos\theta +a\theta sin\theta$

$\begin{array}{l}\therefore \frac{dx}{d\theta }=-asin\theta +asin\theta +a\theta cos\theta =a\theta cos\theta \\ y=asin\theta -a\theta cos\theta \\ \therefore \frac{dy}{d\theta }=acos\theta -acos\theta +a\theta sin\theta =a\theta sin\theta \\ \therefore \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\frac{a\theta sin\theta }{a\theta cos\theta }=tan\theta \end{array}$

$\therefore$ Slope of the normal at any point $\theta$ is $-\frac{1}{tan\theta }$

The equation of the normal at a given point $\left(x,y\right)$ is given by,

$\begin{array}{l}y-asin\theta +a\theta cos\theta =\frac{-1}{tan\theta }\left(x-acos\theta -a\theta sin\theta \right)\\ \Rightarrow ysin\theta -asi{n}^2\theta +a\theta sin\theta cos\theta =-xcos\theta +aco{s}^2\theta +a\theta sin\theta cos\theta \\ \Rightarrow xcos\theta +ysin\theta -a\left(si{n}^2\theta +co{s}^2\theta \right)=0\\ \Rightarrow xcos\theta +ysin\theta -a=0\end{array}$

Now, the perpendicular distance of the normal from the origin is

<p>We have&nbsp;<span title="Click to copy mathml"><math><mrow><mi>x</mi><mo>=</mo><mi>a</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi></mrow></math></span></p><p><span title="Click to copy mathml"><math><mtable columnalign="left"><mtr><mtd><mrow><mo>∴</mo><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>θ</mi></mrow></mfrac><mo>=</mo><mo>−</mo><mi>a</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>=</mo><mi>a</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi></mrow></mtd></mtr><mtr><mtd><mrow><mi>y</mi><mo>=</mo><mi>a</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi></mrow></mtd></mtr><mtr><mtd><mrow><mo>∴</mo><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>θ</mi></mrow></mfrac><mo>=</mo><mi>a</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>=</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi></mrow></mtd></mtr><mtr><mtd><mrow><mo>∴</mo><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>θ</mi></mrow></mfrac><mo>.</mo><mfrac><mrow><mi>d</mi><mi>θ</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi></mrow><mrow><mi>a</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi></mrow></mfrac><mo>=</mo><mi>t</mi><mi>a</mi><mi>n</mi><mi>θ</mi></mrow></mtd></mtr></mtable></math></span></p><p><span title="Click to copy mathml"><math><mrow><mo>∴</mo></mrow></math></span>&nbsp;Slope of the normal at any point&nbsp;<span title="Click to copy mathml"><math><mrow><mi>θ</mi></mrow></math></span>&nbsp;is&nbsp;<span title="Click to copy mathml"><math><mrow><mo>−</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>t</mi><mi>a</mi><mi>n</mi><mi>θ</mi></mrow></mfrac></mrow></math></span></p><p>The equation of the normal at a given point&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>,</mo><mi>y</mi></mrow><mo>)</mo></mrow></mrow></math></span>&nbsp;is given by,</p><p><span title="Click to copy mathml"><math><mtable columnalign="left"><mtr><mtd><mrow><mi>y</mi><mo>−</mo><mi>a</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>=</mo><mfrac><mrow><mo>−</mo><mn>1</mn></mrow><mrow><mi>t</mi><mi>a</mi><mi>n</mi><mi>θ</mi></mrow></mfrac><mrow><mo>(</mo><mrow><mi>x</mi><mo>−</mo><mi>a</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi></mrow><mo>)</mo></mrow></mrow></mtd></mtr><mtr><mtd><mrow><mo>⇒</mo><mi>y</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>=</mo><mo>−</mo><mi>x</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>+</mo><mi>a</mi><mi>c</mi><mi>o</mi><msup><mrow><mi>s</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>θ</mi><mo>+</mo><mi>a</mi><mi>θ</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi></mrow></mtd></mtr><mtr><mtd><mrow><mo>⇒</mo><mi>x</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>+</mo><mi>y</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mrow><mo>(</mo><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>θ</mi><mo>+</mo><mi>c</mi><mi>o</mi><msup><mrow><mi>s</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>θ</mi></mrow><mo>)</mo></mrow><mo>=</mo><mn>0</mn></mrow></mtd></mtr><mtr><mtd><mrow><mo>⇒</mo><mi>x</mi><mi>c</mi><mi>o</mi><mi>s</mi><mi>θ</mi><mo>+</mo><mi>y</mi><mi>s</mi><mi>i</mi><mi>n</mi><mi>θ</mi><mo>−</mo><mi>a</mi><mo>=</mo><mn>0</mn></mrow></mtd></mtr></mtable></math></span></p><p>Now, the perpendicular distance of the normal from the origin is</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1734087410phpZY1Fs7_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1734087410phpZY1Fs7.jpeg" alt="" width="396" height="70"></picture></div></div>

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