Question

Prove that the line 2x - 3y = 9 touches the conics y² = -8x. Also, find the point of contact.

Answer 1

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)`