The length of the perpendicular from the origin,
on the normal to the curve,
x2 + 2xy – 3y2 = 0
at the point (2,2) is
Solution
x<sup>2</sup> + 2xy – 3y<sup>2</sup> = 0
<br><br>Differentiate the curve
<br><br> 2x + 2y + 2xy' – 6yy' = 0
<br><br>$\Rightarrow$ $\Rightarrow$ x + y + xy' – 3yy' = 0
<br><br>$\Rightarrow$ y'(x – 3y) = – (x + y)
<br><br>$\Rightarrow$ y' = ${{x + y} \over {3y - x}}$
<br><br>Slope of normal = $- {{dx} \over {dy}}$ = ${{x - 3y} \over {x + y}}$
<br><br>$\therefore$ ${\left( { - {{dx} \over {dy}}} \right)_{\left( {2,2} \right)}}$ = ${{2 - 6} \over {2 + 2}}$ = -1
<br><br> Normal at (2, 2)
<br><br>y – 2 = – 1 (x – 2)
<br><br>$\Rightarrow$ y + x = 4
<br><br>$\therefore$ Perpendicular distance from (0,0)
<br><br>= $\left| {{{0 + 0 - 4} \over {\sqrt 2 }}} \right|$ = ${2\sqrt 2 }$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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