EQUATION OF TANGENT THAT IS PARALLEL OR PERPENDICULAR TO THE LINE

Find the equations of the tangents.

Problem 1 :

To the parabola y2 = 6x, parallel to 3x - 2y + 5 = 0.

Solution :

y2 = 6x

y2 = 4ax

6x = 4ax

6/4 = a

3/2 = a

Equation of tangent y = mx + a/m

Slope of the tangent = Slope of the line

= -(a/b)

= -(3/-2)

m = 3/2

y = 32x + 3232= 32x + 32 × 23y = 32x + 1y = 3x + 222y = 3x + 23x - 2y + 2 = 0

So, equation of tangents is 3x - 2y + 2 = 0.

Problem 2 :

To the parabola y2 = 16x, perpendicular to the line 3x - y + 8 = 0.

Solution :

y2 = 16x

y2  = 4ax

16x = 4ax

16/4 = a

4 = a

Equation of tangent y = mx + a/m

Slope of the tangent = Slope of the line

= b/a

m = -1/3

y = -13x + 4-13= -13x + 4 × -3y = -13x - 12y = -x - 3633y = -x - 36x + 3y + 36 = 0

So, equation of tangents is x + 3y + 36 = 0.

Problem 3 :

To the ellipse x220 + y25 = 1, which are perpendicular to x + y + 2 = 0.

Solution :

x220 + y25 = 1a2 = 20, b2 = 5

Slope of the tangent = Slope of the line

= b/a

m = 1/1

m = 1

Equation of tangents is y = mx ± a2m2 + b2y =x ± 20(1)2 + 5y = x ± 25y = x ± 5

So, equation of tangents is x - y + 5 = 0.

Problem 4 :

To the hyperbola 4x2 - y2 = 64, which are parallel to 10x - 3y + 9 = 0.

Solution :

4x2 - y2 = 64

Equation of tangent y = mx + a/m

Slope of the tangent = Slope of the line

= -(a/b)

= -(10/-3)

m = 10/3

x2a2 - y2b2 = 14x264 - y264 = 6464x216 - y264 = 1 a2 = 16, b2 = 64Equation of tangents is y = mx ± a2m2 - b2y =103x ± 161032 - 64y =103x ± 161009 - 64y =103x ± 1600 - 5769y =103x ± 10249y =103x ± 3233y = 10x + 3210x -3y + 32 = 0

So, equation of tangents is 10x - 3y + 32 = 0.

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