FIND THE EQUATION OF TANGENT TO THE CURVE IT IS PARALLEL TO THE LINE

Example 1 :

Find the equations of tangents to the hyperbola (x²/16) - (y²/64)  =  1 which is parallel to 10x - 3y + 9  =  0.

Solution :

Slope of the tangent line drawn to the hyperbola is parallel to the line 10x - 3y + 9  =  0

  m  =  -coefficient of x/coefficient of y 

  m  =  -10/(-3)  =  1 0/3

a²  =  16 and b²  =  64

y = mx ± √(a²m² - b²)  ---(1)

y = (10/3) x ± √(16(100/9)) - 64)

y = (10/3) x ± √(1600/9) - 64

y = (10/3) x ± √(1600 - 576)/9

y = (10/3) x ± (32/3)

3y = 10x ± 32

10x - 3y + 32  =  0, 10x - 3y - 32  =  0

Hence the required equations are 10x - 3y + 32  =  0, 10x - 3y - 32  =  0.

Example 2 :

Show that the line x − y + 4 = 0 is a tangent to the ellipse x² + 3y² = 12 . Also find the coordinates of the point of contact.

Solution :

Condition for tangency to the ellipse.

c²  =  a²m² + b²   ----(1)

x² + 3y² = 12 

(x²/12) + (y²/4) = 1

a²  =  12 and b²  =  4

x − y + 4 = 0

  m = -coefficient of x/coefficient of y

  m  =  -1/(-1)    =  1

c  =  4

4²  =  12(1) + 4

16  =  16

Hence the given line is the tangent for the ellipse.

Example 3 :

Find the equation of the tangent to the parabola y² = 16x perpendicular to 2x + 2y + 3 = 0 .

Solution :

y² = 16x

4a  =  16 

a  =  4

The required tangent line is perpendicular to the line 2x + 2y + 3  =  0.

m  =  -2/2  =  -1

Slope of the required line  =  -1/m  =  -1/(-1)  =  1

Equation of the required tangent line :

y  =  mx + (a/m)

y  =  1(x) + (4/1)

y  =  x + 4

x - y + 4  =  0

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