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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 12 Maths Textbook Solution.

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ANSWER: Equation of tangent, \frac{x x_{1}}{a_{2}}+\frac{y y_{1}}{a_{2}}=1

                 Equation  of normal,  \frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}


 Differentiating the given equation


\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { at }\left(x_{1}, y_{1}\right)                                        .....(1)


Since P\left(x_{1}, y_{1}\right) lies on the curve(i)

\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1                                        .......(ii)

\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0

\Rightarrow \frac{d y}{d x}=-\frac{b^{2} x_{1}}{a^{2} y_{1}}

Equation of tangent at P\left(x_{1}, y_{1}\right)is

\begin{aligned} &\left(y-y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}\left(x-x_{1}\right) \\ &\Rightarrow\left(y-y_{1}\right)=-\frac{b^{2} x_{1}}{a^{2} y_{1}}\left(x-x_{1}\right) \\ &\Rightarrow \frac{y y_{1}-y_{1}^{2}}{b^{2}}=\left(\frac{-x x_{1}+x_{1}^{2}}{a^{2}}\right) \end{aligned}

\Rightarrow \frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1        \left [ Using(i) \right ]

Equation of Normal  at P\left(x_{1}, y_{1}\right) is,

\Rightarrow y-y_{1}=-\frac{1}{\left[\frac{d y}{d x}\right]_{\left.\mid x_{1}, y_{1}\right\}}}\left(x-x_{1}\right)

\Rightarrow \frac{b^{2}\left(y-y_{1}\right)}{y_{1}}=\frac{a^{2}\left(x-x_{1}\right)}{x_{1}}

\Rightarrow \frac{b^{2} y}{y_{1}}-b^{2}=\frac{a^{2} x}{x_{1}}-a^{2}

\Rightarrow \frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}                   \left [ Using\left ( ii \right ) \right ]


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