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Question
Prove that the line 2x - 3y = 9 touches the conics y2 = -8x. Also, find the point of contact.
Solution
y2 = - 8x ...(i)
y2 = - 4ax
⇒ a = 2
2x - 3y = 9 ...(ii)
⇒ `(2"x")/3 - 9/3 = "y"` ⇒ y = `2/3` x - 3
m = `2/3` , c = -3
Substituting (ii) in (i) we get :
⇒ `((2"x"-9)/3)^2` = - 8x
⇒ `(4"x"^2 + 81 -36"x")/9 = -8"x"`
⇒ 4x2 + 81 - 36x = -72x
⇒ 4x2 - 36x + 72x +81 = 0
⇒ 4x2 + 36x + 81 = 0
⇒ x = `(-36 ± sqrt((36)^2 - 4xx4xx81))/8`
x = `(-36 ± 0)/8`
x = `(-36)/8 = -9/2`
⇒ x = `-9/2, -9/2`
So the line (ii) meet the parabola (i) at two co-incident points, hence line (ii) meet the parabola (i)
y = `2/3x - 3 = 2/3((-9)/2) -3`
= - 3 - 3 = -6
∴ Point of contact is `(-9/2,-6)`
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