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).

y = 5x2 + 5452= 5x2 + 54 × 25y = 5x2 + 122y = 5x + 1

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

y = x10 + 54110= x10 + 54 × 101y = x10 + 252y = x10 + 1251010y = x + 125
The points of contact are given by am2, 2amWher a = 54, m = 52 and 110If a = 54, m = 52= 54522, 25452= 54254, 10452= 54 × 425, 104 × 25 = 15, 1If a = 54, m = 110= 541102, 254110= 541100, 104110= 54 × 1001, 104 × 101 = (125, 25)

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 

x2a2 - y2b2 = 1x2 - 9y2 = 9Dividing 9 on each sides.x29 - 9y29 = 99x29 - y21 = 1

a2 = 9

a = 3

b2 = 1

b = 1

c2 = a2m2 - b2

9122 = 9-5122 - 181144 = 925144 - 181144 = 225144 - 181144 = 225 - 14414481144 = 81144Point of contact = -a2mc, -b2c= -9 × -5 × 1212 × 9, -1 × 129= 5, -43

∴ 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.

x212 + 3y212 = 1212x212 + y24 = 1 --- (1)x2a2 + y2b2 = 1a2 = 12, y2 = 4

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 = -a2mc, b2c= -12 × 14, 44= -3, 1

∴ point of contact is (-3, 1).

Recent Articles

  1. Solving Word Problems on Numbers Worksheet

    Feb 25, 24 07:44 AM

    Solving Word Problems on Numbers Worksheet

    Read More

  2. Solving Word Problems on Numbers

    Feb 24, 24 11:07 PM

    Solving Word Problems on Numbers

    Read More

  3. Reducing Ratios to Lowest Terms Worksheet

    Feb 24, 24 08:49 PM

    Reducing Ratios to Lowest Terms Worksheet

    Read More