CONDITION FOR TANGENCY TO PARABOLA ELLIPSE AND HYPERBOLA

Example 1 :

Prove that the line 5x-2y+1  =  0 touches the parabola

y²  =  5x

Also find the point of contact.

Solution :

Condition for tangency to parabola :

c  =  a/m  ---(1)

To find "a" :

4a  =  5

a  =  5/4

Equation of the line : 

5x-2y+1  =  0

By applying the values of a, c and m in (1), we get

1/2  =  (5/4)/(5/2)

1/2  =  1/2

Point of Contact :

(a/m², 2a/m) is the point of contact of parabola.

So, the point of contact is (4/5, 1).

Example 2 :

Prove that the line 5x + 12y = 9 touches the hyperbola

x² − 9y² = 9

Solution :

Condition for tangency to hyperbola  :

c²  =  a²m²+b²  ------(1)

Converting into standard form, we get

x² − 9y² = 9

Dividing by 9, we get

x²/9 − y² = 1

a² =  9, b²  =  1

Equation of the line :

5x + 12y = 9

12y  =  -5x+9

y  =  (-5x/12) + (9/12)

y  =  (-5x/12) + (3/4)

Slope (m)  =  -5/12 and y-intercept(c)  =  3/4

By applying the values of a, b, c and m in (1), we get

c²  =  a²m²-b² 

(3/4)²  =  9(-5/12)² - 1²

9/16  =  9(25/144)-1

9/16  =  (25/16)-1

9/16  =  9/16

Example 3 :

Show that the line x − y + 4 = 0 is a tangent to the ellipse

x² + 3y² = 12

Find the coordinates of the point of contact.

Solution :

Condition for tangency to ellipse  :

c²  =  a²m²+b²  ------(1)

Converting into standard form, we get

x² + 3y² = 12

Dividing by 12, we get

x²/12 + y²/4 = 1

a² =  12, b²  =  4

Equation of the line :

 x − y + 4 = 0 

y  =  x+4

Slope (m)  =  1 and y-intercept(c)  =  4

By applying the values of a, b, c and m in (1), we get

c²  =  a²m²+b² 

4²  =  12(1)² + 4

16  =  12+4

16  =  16

So, the given is the tangent line of ellipse.

Point of contact :

(-a²m/c, b²/c) is the point of contact of ellipse.

So, the point of contact is (3, 1).

Example 4 :

The circle x² + y² = 4 is concentric with the ellipse x²/7 + y²/3 = 1; prove that the common tangent is inclined to the major axis at an angle 30^o and find its length.

Solution:

The ellipse x²/7 + y²/3 = 1 ...... (1)

The equation to the circle is

x² + y² = 4 ...... (2)

As the line y = mx + √(a²m² + b²)

i.e. y = mx  + √(7m² + 3) ...... (3)

is always the tangent on the ellipse. If this is also a tangent on the circle (2) then length of perpendicular from the centre (0, 0) on the line (1) must be equal to the radius of circle i.e. 2.

Hence, √(7m²  + 3)/√(1+ m²) = 2

⇒ 7m² + 3 = 2²(1 + m²)

⇒ 7m² - 4m² = 4 - 3

⇒ m² = 1/3

⇒ m = + 1/√3

Hence the common tangent to the two curves is inclined at an angle of tan^-1 (+1/√3) i.e. 30^o to the axis.

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