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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15 .2 Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 1 Maths Textbook Solution.

Answers (1)

Answer:x+y=\frac{a^{2}}{2}

HINTS:

We find the slope and differentiate slope

\Rightarrow \frac{d y}{d x}=-1

Given:

\sqrt{x}+\sqrt{y}=a \text { at the point }\left(\frac{a^{2}}{4}, \frac{a^{2}}{4}\right)

Solution:

\begin{aligned} &\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=0 \\ &\Rightarrow \frac{1}{2 \sqrt{\frac{a^{2}}{4}}}+\frac{1}{2 \sqrt{\frac{a^{2}}{4}}} \cdot \frac{d y}{d x}=0 \\ &\Rightarrow \frac{1}{a}+\frac{1}{a} \cdot \frac{d y}{d x}=0 \\ &\Rightarrow \frac{1}{a} \cdot \frac{d y}{d x}=-\frac{1}{a} \\ &\frac{d y}{d x}=-1 \end{aligned}

\begin{aligned} &\text { equation of tangent will be }\left(y-y_{1}\right)=m\left(x-x_{1}\right)\\ &y-\frac{a^{2}}{4}=-\left(x-\frac{a^{2}}{4}\right)\\ &x+y=\frac{a^{2}}{2} \end{aligned}

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