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

Prove that, the normal to y2 = 12x at (3, 6) meets the parabola again in (27, −18) & circle on this normal chord as diameter is x2 + y2 − 30x + 12y − 27 = 0.

Prove that, the normal to y2
 = 12x at (3,
 6) meets the parabola again in (27, −18) & circle on this normal
chord as diameter is x2
 + y2 − 30x + 12y − 27 = 0.

Grade:

1 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
6 years ago
Equation of tangent to parabola y2=4ax at (x1,y1) is
y-y1 = -y1 (x-x1)/2a
Equation of normal at (3,6) for above parabola will be
y- 6 = -6( x- 3)/(2*3)
On simplifying
x+ y =9
Now solve it with parabola by putting x=9-y in equation of parabola and find other root.
Now we have end points of diameter (3,6) and ( 27,-18)
equation of circle with (x1,y1) and (x2,y2) as diameter is
( x-x1)(x-x2) + (y-y1)(y-y2)=0
Put the values and get the required answer.

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