Click to Chat

1800-2000-838

+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
`        A curve that passes through (2, 4) and having subnormal of constant length of 8 units can be?`
7 years ago

Kevin Nash
332 Points
```										Dear Rishi,
Let the curve be y = f(x). Subnormal at any point = |y*dy/dx|
y*dy/dx = ±8  ;which indicates y dy = ±8dx ;which indicates  = ±8x + c
which indicates: y2 = 16 x+2c1,
: c1= -8    or  y2 = -16x +2c2,
:c2= 24
Hence :        y2 = 16x – 8; y2 = -16x + 24  ;
would be the desired equations.
askIITians.com provides online iit jee courses and IIT JEE Test Series with IITians. Click here to get free online test series and check your status timely or you can join us as our registered user for getting best iit jee study material or iit jee test series.
for further details visit the folowing links:
Askiitians provide you all the best possible information, feel free to ask.
all the best.
thanks and regards.
Akhilesh Shukla
```
7 years ago
18 Points
```										length of subnormal =y(dy/dx)
=8 ......(given)
y(dy/dx)=8
y dy = 8 dx
integrating on both sides
y^2 =16x +c (c is integrating constant)
satisfying by (2,4)
we get y^2 = 16(x-1)
this is the equation of required curve.
From,

```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details