#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# The shortest distance between the line y-x=1 and curve x=y^2 is?Please give detailed answer and explanationThanks 8 years ago

let us take a point on line y - x = 1 : A ( t -1 , t ) and on curve x = y2 : B ( t2 , t )

AB = root [ ( t - t )2 + (t2 - t + 1)2 ] = t2 - t + 1

f(x) = t2 - t + 1 , f '' (x) = 2t - 1 . For min value f '' (x) = 0 ; 2t - 1 =0 ; t =1/2

f '' '' (x) = 2 > 0 therefore t=1/2 is point of local minima

thus shortest distance = (1/2)- (1/2) + 1 = 3/4 units

8 years ago

Any point on parabola, (k2,k)

Perpendicular distance formula:

D=(k-k2-1)/21/2

Differentiating and putting =0

1-2k=0

k=1/2

Therefore the point is (1/4, 1/2)

D=3/(321/2)

8 years ago

But sir can you just tell me how that perpendicular distance formula is used. I would be really grateful to you.

Thanx once again

8 years ago

Distance of any point(x1,y1) from a line ax+by+c is:

|ax1 + by1 + c|/(a2+b2)1/2

That is the formula. You can reply to this post if you want the derivation.

4 years ago

Distance of any point(x1,y1) from a line ax+by+c is:

|ax1 + by1 + c|/(a2+b2)1/2

That is the formula. You can reply to this post if you want the derivation.
4 years ago
why cant we use the method used by bhaveen kumar...can i please know what is the mistake in it and why arent the answers matching???
4 years ago
Mannon Mani, we cannot use that nethod because he took the points in both the curve and line with respect to t, which is incorrect as they are non intersecting. They cannot be taken with the same parameter
4 years ago
what if i use the distance formulae and ill get s= mag( t-t^2-1/ sqrt 1+1)
=> s= t^2-t+1/sqrt(2) { since t^2-t+1 = (t-1/2) ^2+3 /4 >0}
and then ds/dt = 1/sqrt2 (2t-1)
&d^2 s / dt ^ 2= sqrt(2) >0
now ds/dt=0 => t=1/2
s is minimum at t=1/2 and the req shortest dist is(1/2)^2 – (1/2) +1 / sqrt(2)=3/4sqrt(2) => 3sqrt(2)/8
3 years ago
Let (a2,a) be thw point of shortest distance on x=y2The distance between (a2,a) and line x−y+1=0 is given byD=a2−a+12–√=12–√[(a−12)2+34]If is min when a=12 and Dmin=342–√=>32–√8
9 months ago
Dear Student,

Any point on parabola, (k^2,k)
Perpendicular distance formula:
D = (k-k^2-1)/2^(1/2)
Differentiating and putting =0
1-2k = 0
k = 1/2
Therefore the point is (1/4, 1/2)
D = 3/(32^1/2)

Thanks and Regards