Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 19 Sub Question 2 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 19 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer: \text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(3,0) \&(-3,0)

Hint: \text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }

Given: \frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \quad \rightarrow(1)

Solution: \text { Differentiating eqn(1) with respect to } x

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &\frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0 \\ &\frac{d y}{d x}=\frac{-16}{9} \frac{x}{y} \end{aligned}

\text { Now the tangent is parallel to the } \mathrm{y} \text { -axis if the slope of the normal is zero }

\text { slope of the normal }=\frac{-1}{\left(\frac{-16 x}{9 y}\right)} \Rightarrow \frac{9 y}{16 x}=0 \text { which is possible if } y=0

Then,

\frac{x^{2}}{9}+\frac{y^{2}}{16}=1 \text { for } \mathrm{y}=0

x^{2}=9

x=\pm 3

\text { Thus the points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(3,0) \&(-3,0)

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2027 (OUT) LIVE: Class 10, 12 exam dates out for over 46 lakh students; passing marks
anam-khan

CBSE Class 12 exams clash with Holi on March 22; will board revise date sheet?
anam-khan

CBSE Class 12 West Asia result 2026 out; check marks on DigiLocker

Latest Question