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  • Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 18 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 18 maths Textbook Solution.

Answers (1)

Answer: To Prove \frac{x}{a}+\frac{y}{b}=2(R.H.S=L.H.S)

Hint: Differentiate

Given:\left [ \frac{x}{a} \right ]^{n}+\left [ \frac{y}{b} \right ]^{n}=2

Solution:\left [ \frac{x}{a} \right ]^{n}+\left [ \frac{y}{b} \right ]^{n}=2

\begin{aligned} &\Rightarrow \frac{n}{a}\left[\frac{x}{a}\right]^{n-1}+\frac{n}{b}\left[\frac{y}{b}\right]^{n-1} \frac{d y}{d x}=0 \\ &\Rightarrow \frac{n}{b}\left[\frac{y}{b}\right]^{n-1} \frac{d y}{d x}=-\frac{n}{a}\left[\frac{x}{a}\right]^{n-1} \\ &\Rightarrow \frac{d y}{d x}=-\frac{n}{a}\left[\frac{x}{a}\right]^{n-1} \times \frac{b}{n}\left[\frac{b}{y}\right]^{n-1}=-\frac{b}{a}\left(\frac{b x}{a y}\right)^{n-1} \end{aligned}

\therefore slope \: of \, tangent=\left(\frac{d y}{d x}\right)_{(a, b)}=-\frac{b}{a}\left(\frac{b \times a}{a \times b}\right)^{n-1}=-\frac{b}{a}....(ii)

The equation of the tangent is

\begin{aligned} &y-b=-\frac{b}{a}(x-a) \\ &\Rightarrow y a-a b=-x b+a b \\ &\Rightarrow x b+y a=2 a b \\ &\Rightarrow \frac{x}{a}+\frac{y}{b}=2 \end{aligned}

So the given line touched the given curve at the given point

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