Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
846 views
in Calculus by (114k points)
closed by
Find the equation of the normal to the curve 2x2 = y, which passes through (1, 2)
1. x + y + 9 = 0
2. 4x + y - 9 =0
3. 3x + 4y - 8 = 0
4. x + 4y - 9 = 0 
5. None of these

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 4 : x + 4y - 9 = 0 

Concept:

Slope of tangent to the curve = \(\rm \frac {dy}{dx}\)

Slope of normal to the curve = \(\rm \frac{-1}{\left(\frac {dy}{dx}\right)}\)

Point-slope is the general form: y - y1 = m(x - x1), Where m = slope

Calculation:

Here, y = 2x2

\(\rm \frac {dy}{dx}\) = 4x

\(\rm \left.\frac{dy}{dx}\right|_{x=1}=4\)

Slope of normal to the curve =\(\rm \frac{-1}{\left(\frac {dy}{dx}\right)}\) = -1/4

Equation of normal to curve passing through (1, 2) is y - 2 = -1/4(x - 1)

⇒ 4y - 8 = -x + 1

⇒ x + 4y - 9 = 0 

Hence, option (4) is correct.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...