IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nadeemoidu

Find the shortest distance of the point (0,c) from the parabola y = x^2

where

.
(Part 1 - Pg 225)

The answer given is rac{sqrt{4c-1}}{2}

which is obviously wrong for c < 1/4.

I'm getting the shortest distance = c for c< 1/4 .

Isn't it?

junglejalebi

it is correct man .
recheck it.

nadeemoidu

Then what do u get for c= 0?

junglejalebi

brother,
u haven,t understood the question.
it is simple yaar
thode thande dimaag se sooch.
dont panic

junglejalebi

u are right man.
c=0 par to doesnt make any sense.

sboosy

Hi nadeem ..ur spot on ..

distance is

(x²+(x²-c)²)

the differenciated distance eqn is

2x(1+2(x²-c) ) =0

so x = 0 or x² - c = -1/2

x² = c-1/2

but this as we see is valid only for c>=1/2

but since c can be below 1/2 upto 0 there shud a splitting

so distance is

(4c-1) / 2 for c>=1/2

else we take the case that x is 0

which gives c

konichiwa2x

what made you do ncert examples?

nadeemoidu

@konichiwa

Board exams!!

After all i'm studying in a school where less than 10 of the total 400 class 12 science students r appearing for the JEE.

hsbhatt

. But I liked the advice given by that newbie to nadeem to relax. He should have atleast noted that he has notched up 773 points so he is no novice!