Get Answers to all your Questions

header-bg qa

Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 19 maths Textbook Solution.

Answers (1)

Answer: Equation of tangent, \sqrt{2} b x-a y-a b=0

             Equation of normal, a x+\sqrt{2} b y-\sqrt{2}\left(a^{2}+b^{2}\right)=0

Hint: Differentiating the given curve with respect to x.

Given\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\: at\left ( \sqrt{2a}, b \right )

Solution: \frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}

\left ( \frac{dy}{dx} \right )_{\sqrt{2a,b}}=\frac{\sqrt{2b}}{a}

Slope of tangent is

\frac{\sqrt{2b}}{a}and of the normal is

The equation of tangent is

\begin{aligned} &y-b=\frac{\sqrt{2 b}}{a}(x-\sqrt{2} a) \\ &\sqrt{2} b x-a y-a b=0 \end{aligned}

Equation of normal is,

\begin{aligned} &y-b=\frac{-a}{\sqrt{2} b}(x-\sqrt{2} a) \\ &a x+\sqrt{2} b y-\sqrt{2}\left(a^{2}+b^{2}\right)=0 \end{aligned}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support