Thank you for registering.

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

Please check your email for login details.
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 circle cuts the parabola y2 = 4ax in the points ( ati2, 2ati) for i = 1, 2, 3, 4. Prove that ti + t2 + t3 + t4 = O.

A circle cuts the parabola y2   = 4ax in the points ( ati2, 2ati) for i = 1, 2, 3, 4. Prove that ti   + t2   + t3   + t4   = O. 

Question Image
Grade:12

1 Answers

Arun
25763 Points
2 years ago

We can proceed in the following way

Let the equation of the circle be x^2+y^2+2*g*x+2*f*y+c=0

Let the equation of the parabola be y^2=4ax.

Now replace x^2 by y^4/16a^2 in the equation of circle and x by y^2/4a.(To find the ordinates of the intersection points).

Now the equation of the circle becomes.

y^4/16*a^2+y^2+2*g*(y^2/4a)+2*f*y+c=0 (The roots of the equation give the value of the ordinates of the point of intersection)

The co-efficient of y^3 is 0. … (In general).(The roots may be real or complex).

So Sum of ordinates is 0.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free