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
+1 vote
505 views
in Derivatives by (28.3k points)
closed by

Find the point on the curve y2 = 2x which is at a minimum distance from the point (1, 4).

1 Answer

+1 vote
by (29.4k points)
selected by
 
Best answer

Given Curve is y2 = 2x …… (1) 

Let us assume the point on the curve which is nearest to the point (1, 4) be (x, y) 

The (x, y) satisfies the relation(1) 

Let us find the distance(S) between the points (x, y) and (1,4) 

We know that distance between two points (x1,y1) and (x2,y2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

 ⇒ S = \(\sqrt{(x-1)^2+(y-4)^2}\)

⇒ S = \(\sqrt{x^2-2x+1+y^2-8y+16}\)

⇒ S = \(\sqrt{x^2-2x+y^2-8y+17}\)

Squaring on both sides we get, 

⇒ S2 = x2 + y2 - 2x - 8y + 17 

From (1)

We know that distance is an positive number so, for a minimum distance S, S2 will also be minimum 

Let us S2 as the function of y. 

For maxima and minima,

We get minimum distance at y = 2 

Let find the value of x at these y values 

⇒ x = \(\frac{(2)^2}{2}\)

⇒ x = 2 

∴ The nearest point to the point (2, 2) on the curve is (4,4).

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

...