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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 29 of 50
Marks: +1, -0
The equation of the normal to the curve ay2=x3a y^2=x^3 at the point (am2,am3)(a m^2, a m^3) is
Solution:     👈: Video Solution
Given equation of curve is
a2=x3a^2=x^3
Differentiating w.r.t. xx, we get
   2ay   dydx=3x2\;\text{ 2ay }\;\frac{dy}{dx}=3x^2
⇒  dydx=  3x22ay\Rightarrow \;\frac{dy}{dx}=\;\frac{3x^2}{2ay}
Slope of the tangent to the curve at (am2,am3)(am^2, am^3) is
(  dydx)(am2,am3)=  3(am2)22a(am3)=  3a2m42a2m3=  3m2\left(\;\frac{dy}{dx}\right)_{(a m^2, a m^3)} =\;\frac{3 (a m^2)^2}{2 a (a m^3)} =\;\frac{3 a^2 m^4}{2 a^2 m^3} =\;\frac{3m}{2}
Slope of normal at (am2,am3)(am^2, am^3)
=  −1   slope of the tangent at   (am2,am3)=  −23m=\;\frac{-1}{\;\text{ slope of the tangent at }\;(am^2, am^3)} =\;\frac{-2}{3m}
Equation of the normal at (am2,am3)(am^2, am^3) is
y−am3=  −23m(x−am2)y - a m^3 = \;\frac{-2}{3m} (x - a m^2)
⇒3my−3am4=−2x+2am2\Rightarrow 3 m y - 3 a m^4 = -2 x + 2 a m^2
⇒2x+3my−am2(2+3m2)=0\Rightarrow 2 x + 3 m y - a m^2 (2 + 3 m^2) = 0
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