Equation of tangent for the parabola from external point :
Equation of tangent for the ellipse from external point :
Equation of tangent for the hyperbola from external point :
Find the equation of the two tangents that can be drawn.
Problem 1 :
From the point (2, -3) to the parabola y^{2} = 4x.
Solution :
Given, y^{2} = 4x
y^{2} = 4ax
4x = 4ax
a = 1
Equation of the tangent to the parabola will be of the form
y = mx + (a/m)
Applying the value of a, we get
y = mx + 1/m
The tangent passes through the point (2, -3)
-3 = 2m + 1/m
-3 = (2m^{2} + 1)/m
-3m = 2m^{2} + 1
2m^{2} - 3m + 1 = 0
So, the equation of the two tangents is
x + 2y + 4 = 0; x + y + 1 = 0.
Problem 2 :
From the point (1, 3) to the ellipse 4x^{2} + 9y^{2} = 36.
Solution :
Given, 4x^{2} + 9y^{2} = 36
Dividing 36 on each sides.
Equation of tangent :
So, the equation of the two tangents is x - 2y + 5 = 0; 5x + 4y - 17 = 0.
Problem 3 :
From the point (1, 2) to the hyperbola 2x^{2} - 3y^{2} = 6.
Solution :
Given, 2x^{2} - 3y^{2} = 6
Dividing 6 on each sides.
Equation of tangent :
y - y_{1} = m(x - x_{1})
when m = 1
y - 2 = 1(x - 1)
y - 2 = x - 1
y - 2 - x + 1 = 0
y - x - 1 = 0
Dividing -1 on each sides.
x - y + 1 = 0
when m = -3
y - 2 = -3(x - 1)
y - 2 = -3x + 3
y - 2 + 3x - 3 = 0
3x + y - 5 = 0
So, the equation of the two tangents is 3x + y - 5 = 0; x - y + 1 = 0.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM