FIND EQUATIONS OF TANGENT DRAWN FROM THE POINT AWAY FROM THE CURVE
Equation of tangent drawn to the parabola will be in the form :
y = mx + (a/m)
Equation of tangent drawn to the ellipse will be in the form :
y = mx ± √a^{2} m^{2} + b^{2}
Equation of tangent drawn to the hyperbola will be in the form
y = mx ± √a^{2} m^{2} + b^{2}
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 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).
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 :
Find the equations of the two tangents that can be drawn from the point (5, 2) to the ellipse 2x^{2} + 7y^{2} = 14.
Solution :
Given, 2x^{2} + 7y^{2} = 14
Dividing 14 on each sides.
2x214+7y214=1414x27+y22= 1 x2a2+y2b2= 1
Equation of tangent :
y = mx ±a2m2+ b22 =5m ±7m2+ 22 - 5m =7m2+ 2Squaring on both sides.(2 - 5m)2= 7m2+ 222+ 52m2- 2(2)(5m)= 7m2+ 24 + 25m2- 20m = 7m2+ 24 + 25m2- 20m - 7m2- 2 = 018m2- 20m + 2 = 0Dividing 2 on each sides.9m2- 10m + 1 = 09m2- 9m - 1m + 1 = 09m(m - 1)-1(m - 1)= 0(9m - 1)(m - 1)= 0m =19 and m = 1While applying the values of m, we get y = mx ±a2m2+ b2Where m =19y =19x ±7192+ 2y =19x ±7181+ 2y =19x ±781+ 2y =19x ±7 + 16281y =19x ±16981y =19x ±1399y = x + 13x - 9y + 13 = 0Where m = 1y =x ±7(1)2+ 2y = x ±7 + 2y = x ±9y = x ± 3x - y + 3 = 0 or x - y - 3 = 0
Problem 3 :
Find the equation of the two tangents that can be drawn
(i) From the point (2, -3) to the parabola y^{2} = 4x.
(ii) From the point (1, 3) to the ellipse 4x^{2} + 9y^{2} = 36.
(iii) From the point (1, 2) to the hyperbola 2x^{2} - 3y^{2} = 6.
Solution :
(i) Given, y^{2} = 4x
y^{2} = 4ax
4x = 4ax
a = 1
Passes through the points (2, -3).
Equation of the tangent to the parabola will be of the form y = mx + 1/m.