हिंदी

Find the Equation of the Normal to the Curve Ay2 = X3 At the Point (Am2, Am3) ?

Advertisements
Advertisements

प्रश्न

Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3) ?

Advertisements

उत्तर

\[a y^2 = x^3 \]

\[\text {Differentiating both sides w.r.t.x }, \]

\[2ay \frac{dy}{dx} = 3 x^2 \]

\[ \Rightarrow \frac{dy}{dx} = \frac{3 x^2}{2ay}\]

\[\text { Slope of tangent } = \left( \frac{dy}{dx} \right)_\left( a m^2 , a m^3 \right) =\frac{3 a^2 m^4}{2 a^2 m^3}=\frac{3m}{2}\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( a m^2 , a m^3 \right)\]

\[\text { Equation of normal is },\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - a m^3 = \frac{- 2}{3m} \left( x - a m^2 \right)\]

\[ \Rightarrow 3my - 3a m^4 = - 2x + 2a m^2 \]

\[ \Rightarrow 2x + 3my - a m^2 \left( 2 + 3 m^2 \right) = 0\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Tangents and Normals - Exercise 16.2 [पृष्ठ २८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 15 Tangents and Normals
Exercise 16.2 | Q 7 | पृष्ठ २८
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×