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 y^{2} = 5x from the point (5, 13). Also find the points of contact.
Solution :
The equation of the parabola is y^{2} = 5x.
y^{2} = 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 = (20m^{2} + 5)/4m
52m = 20m^{2} + 5
20m^{2} - 52m + 5 = 0
20m^{2} - 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 x^{2} - 9y^{2} = 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
a^{2} = 9 a = 3 |
b^{2} = 1 b = 1 |
c^{2} = a^{2}m^{2} - b^{2}
∴ point of contact is (5, -4/3).
Problem 3 :
Show that the line x - y + 4 = 0 is a tangent to the ellipse x^{2} + 3y^{2} = 12. Find the co-ordinates of the point of contact.
Solution :
x^{2} + 3y^{2} = 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
c^{2} = a^{2}m^{2} + b^{2}
(4)^{2} = (1^{2})(1) + 4
16 = 12 + 4
16 = 16
∴ point of contact is (-3, 1).
Feb 25, 24 07:44 AM
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