# FIND POINT OF CONTACT OF TANGENT AND PARABOLA ELLIPSE AND HYPERBOLA

## Point of Contact of Parabola Ellipse and Hyperbola

Point of contact of parabola :

Point of contact of Ellipse :

Point of contact of Hyperbola :

Problem 1 :

Find the equations of the tangents to the parabola y2 = 5x from the point (5, 13). Also find the points of contact.

Solution :

The equation of the parabola is y2 = 5x.

y2 = 4ax

5x = 4ax

a = 5/4

Equation of the tangent  y = mx + a/m

y = mx + 5/4m --- (1)

Passes through the points (5, 13).

13 = 5m + 5/4m

13 = (20m2 + 5)/4m

52m = 20m2 + 5

20m2 - 52m + 5 = 0

20m2 - 2m - 50m + 5 = 0

2m(10m - 1) - 5(10m - 1) = 0

(2m - 5) (10m - 1) = 0

2m - 5 = 0 and 10m - 1 = 0

2m = 5 and 10m = 1

m = 5/2 and m = 1/10

m = 5/2 substitute the equation (1).

m = 1/10 substitute the equation (1).

So, the points of contact are (1/5, 1), (125, 25).

Problem 2 :

Prove that the line 5x + 12y = 9 touches the hyperbola x2 - 9y2 = 9 and find its point of contact.

Solution :

5x + 12y = 9

12y = 9 - 5x

y = (9 - 5x)/12

y = mx + c

y = -5x/12 + 9/12

m = -5/12, c =  9/12

 a2 = 9a = 3 b2 = 1b = 1

c2 = a2m2 - b2

∴ point of contact is (5, -4/3).

Problem 3 :

Show that the line x - y + 4 = 0 is a tangent to the ellipse x2 + 3y2 = 12. Find the co-ordinates of the point of contact.

Solution :

x2 + 3y2 = 12

Dividing 12 on each sides.

Equation of the line is x - y + 4 = 0.

y = x + 4

y = mx  + c

where m = 1, c = 4

c2 = a2m2 + b2

(4)2 = (12)(1) + 4

16 = 12 + 4

16 = 16

∴ point of contact is (-3, 1).

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