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  • Need Solution for R D Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15 point 2 Question 6 Maths Textbook Solution

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 6 Maths Textbook Solution.

Answers (1)

ANSWER: The equation of normal x=2

HINTS:

 Differentiating  with respect to  x to get its slope.

GIVEN:

x^{2}+2y^{2}-4x-6y+8=0 \: \: at \: \: x=2

SOLUTION:

Upon differentiation

\begin{aligned} &2 x+\left[4 y\left[\frac{d y}{d x}\right]\right]-4-6\left[\frac{d y}{d x}\right]=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{(4-2 x)}{(4 y-6)} \end{aligned}

Finding the y coordinate by substitute x in the given curve

\Rightarrow 2y^{2}-6y+4=0

\Rightarrow y^{2}-3y+2=0

y=2 or y=1

The normal is perpendicular to tangent, therefore , m_{1,}m_{2}=-1

m\left ( normal \right )at\; x=2\: is\: \frac{1}{0} Which is Undefined

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-1=\frac{-1}{0}(x-2) \\ &\Rightarrow-(x-2)=0 \\ &\Rightarrow x-2=0 \\ &\Rightarrow x=2 \end{aligned}

when y=2

Slope of tangent =\left ( \frac{dy}{dx} \right )\left ( _{2,1} \right )=\frac{0}{-1}=0

Equation of normal is

\begin{aligned} &y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-2=\frac{-1}{0}(x-2) \\ &\Rightarrow-(x-2)=0 \\ &\Rightarrow x-2=0 \\ &\Rightarrow x=2 \end{aligned}

In both cases the equation of normal is x=2

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