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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
9 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.

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