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
Menu
Grade: 12th pass
        
Let ax-y+7=0 be a chord of parabola x^2=28y
meeting it at A and B ,and tangents at A and B meet at C.the locus of circumcentre of tr.ABC is
4 years ago

Answers : (1)

Latika Leekha
askIITians Faculty
165 Points
							Hello student,
Given chord is ax – y + 7 = 0 and the parabola is x2 = 28y.
We first try to find out the points of intersection of the chord with the parabola as it would give us the end points of the chord.
So, substituting the value of y = ax+7 from the equation of the chord into the equation of the parabola we have,
x2 = 28(ax+7)
So, x = 14(a ± √a2+1)
y = x2/28
Let us denote the x and y cordinates as x*x** and y*y**.
Then the slope of the tangents at these two points would be given by
m*m** which would be equal to the value of dy/dx at x*x**.
Hence, m*m** = x*x**/14.
This clearly shows that the product of slopes is -1.
Hence, this means that the two tangents are perpendicular and therefore the triangle is a right angled triangle.
Circumcentre of a right angled traingle is given by the mid-point of the hypotenuse.
So, the coordinates of circumcentre are x$ = 14a
y$ = (x*2 + x**2)/ 28 = 7(2a2+1)
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 53 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Get Extra Rs. 148 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details