Question
The line y = mx + c may touch the parabola y2 = 4a (x + a), if-
- c = am – (a/m)
- c = a/m
- c = – a/m
- c = am + (a/m)
Hint:
replace value of y into the equation of parabola.
The correct answer is: c = am + (a/m)
c= a/m + am
Y= mx + c
Y^2 = 4a(x+a)
=> (mx+c)2= 4a(x+a)
m2x2 +(2mc-4a)x+(c2-4a2)=0
D=0 since only one point of contact is present.
(2mc-4a)2-4m2(c2-4a2)=0
This gives us c= a/m + am
the solution of the two curves gives us the quadratic equation. D=0 gives us the required answer.
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