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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 10 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 10 Maths Textbook Solution.

Answers (1)

ANSWER: Equation of tangent, x \cos ^{3} \theta+y \sin ^{3} \theta=c

                 Equation  of normal , x\sin ^{3} \theta-y \cos ^{2} \theta+2 c \cot (2 \theta)=0

HINTS:

 Differentiating the given curve with respect to x and find its slope.  \sin ^{2} \theta+\cos ^{2} \theta=1

GIVEN:

c^{2}\left(x^{2}+y^{2}\right)=x^{2} y^{2} \ \ \ \ at\left[\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right]

SOLUTION:

\begin{aligned} &\Rightarrow 2 x c^{2}+2 y c^{2} \frac{d y}{d x}=x^{2} 2 y \frac{d y}{d x}+2 x y^{2} \\ &\Rightarrow \frac{d y}{d x}\left(2 y c^{2}-2 x^{2} y\right)=2 x y^{2}-2 x c^{2} \\ &\Rightarrow \frac{d y}{d x}=\frac{x y^{2}-x c^{2}}{y c^{2}-x^{2} y} \end{aligned}

Slope of tangent,

m=\left(\frac{d y}{d x}\right)_{\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)}

      =\frac{\frac{c^{3}}{\cos \theta \sin ^{2} \theta}-\frac{c^{3}}{\cos \theta}}{\frac{c^{3}}{\sin \theta}-\frac{c^{3}}{\cos ^{2} \theta \sin \theta}}

       =\frac{\frac{1-\sin ^{2} \theta}{\cos \theta \sin ^{2} \theta}}{\frac{\cos ^{2} \theta-1}{\cos ^{2} \theta \sin \theta}}

        =\frac{\cos ^{2} \theta}{\cos \theta \sin ^{2} \theta} \times \frac{\cos ^{2} \theta \sin \theta}{-\sin ^{2} \theta}

       =\frac{\cos ^{3} \theta}{-\sin ^{3} \theta}

 Given,\left(x_{1}, y_{1}\right)=\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{c}{\sin \theta}=\frac{-\cos ^{3} \theta}{\sin ^{3} \theta}\left(x-\frac{c}{\cos \theta}\right) \\ &\Rightarrow \frac{y \sin \theta-c}{\sin \theta}=\frac{-\cos ^{3} \theta}{\sin ^{3} \theta}\left(\frac{x \cos \theta-c}{\cos \theta}\right) \\ &\Rightarrow \sin ^{2} \theta(y \sin \theta-c)=-\left(\cos ^{2} \theta x \cos \theta-c\right) \\ &\Rightarrow y \sin ^{3} \theta-c \sin ^{2} \theta=-x \cos ^{3} \theta+c \cos ^{2} \theta \\ &\Rightarrow x \cos ^{3} \theta+y \sin ^{3} \theta=c\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ &\Rightarrow x \cos ^{3} \theta+y \sin ^{3} \theta=c \end{aligned}

Equation of Normal  is,

\begin{aligned} &y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{c}{\sin \theta}=\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\left(x-\frac{c}{\cos \theta}\right) \\ &\Rightarrow \cos ^{3} \theta\left(y-\frac{c}{\sin \theta}\right)=\sin ^{3} \theta\left(x-\frac{c}{\cos \theta}\right) \\ &\Rightarrow y \cos ^{3} \theta-c \frac{\cos ^{3} \theta}{\sin \theta}=x \sin ^{3} \theta-c \frac{\sin ^{3} \theta}{\cos \theta} \end{aligned}

\begin{aligned} &\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=c\left(\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\cos \theta \sin \theta}\right) \\ &\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=c\left[\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}{\cos \theta \sin \theta}\right] \\ &\Rightarrow x\sin ^{3} \theta-y \cos ^{3} \theta=2 c\left[\frac{-\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{2 \cos \theta \sin \theta}\right] \end{aligned}

\begin{aligned} &\Rightarrow x\sin ^{3} \theta-y \cos ^{3} \theta=2 c\left[\frac{-\cos (2 \theta)}{\sin (2 \theta)}\right] \\ &\Rightarrow x\sin ^{3} \theta-y \cos ^{3} \theta=-2 c \cot (2 \theta) \\ &\Rightarrow x\sin ^{3} \theta-y \cos ^{3} \theta+2 c \cot (2 \theta)=0 \end{aligned}

 

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