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
`        If PN is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, the midpoint of PN is a/an `
7 years ago

147 Points
```										Dear Zahid
you have not maintion where is point P and N ,
ia m solving this question asuming P is on the parabola and N is on the asymptotes
let equation of hyperbola is   x2 -y2 =a2
and equation os aymptote is  :    y=x   (slope  m=1)
let N(x1,x1) on the asymptote ,and a  point P(asecθ ,atanθ) on the hyperbola and(h,k) is the mid point of PN
h=(x1+asecθ)/2   or  asecθ =2h-x1  ...............1
k = (x1+atanθ)/2  or  atanθ =2k-x1  ................2

since PN is perpendiculat to  aymptote so slope of PN =-1
(k-x1)/(h-x1)=-1
x1 =(h+k)/2
so put this value in equation 1 and 2
asecθ=(3h-k)/2
atanθ =(3k-h)/2

now square and subtract
a2(sec2θ -tan2θ)  = {(3h-k)/2}2 - {(3k-h)/2}2
a2 = {(3h-k)/2}2 - {(3k-h)/2}2
4a2 =8(h2-k2)
h2-k2=a2/2
Please feel free to post as  many doubts on our discussion forum as you can.If you find any  question Difficult to understand - post it here and we will get you the  answer and detailed  solution very  quickly. We are all IITians  and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians  ExpertsBadiuddin
```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

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