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)

Sandeep Pathak
askIITians Faculty
25 Points
							Let us first solve for end points of the chord by finding points of intersection of chord and parabola:
x^2 = 28(ax+7)\\ \Rightarrow x_{\pm} = 14(a\pm\sqrt{a^2+1}), y_{\pm} = \frac{x_{\pm}^2}{28}
It is also interesting to notex_{-}=-x^{-1}_{+}
Let’s now call the slope of tangents at these two points –m_{\pm}
m_{\pm}=\frac{dy}{dx}\Big|_{x_{\pm}}=\frac{x_{\pm}}{14}
So it is easy to see that
m_+\times m_- = -1

Thus, the two tangents are perpendicular and tr.ABC is a right-angled triangle.
AND

Circumcenter of a right-angled triangle is the midpoint of the hypotenuse.

This gives, for the coordinates of circumcenter,
x_o=14a\\ y_o=\frac{1}{28}\left(x_+^2+x_-^2 \right )=7(2a^2+1)
Thus, the locus is a parabola given in parametric form above.

4 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