FIND THE POINTS ON THE CURVE WHERE THE TANGENT IS PARALLEL TO THE LINE

Example 1 :

Find at what points on a circle

x² + y² = 13

the tangent is parallel to the line 2x + 3y = 7.

Solution :

Since the tangent line drawn at the point to the circle is parallel to the given line, their slopes will be equal. 

Slope of the given line 2x + 3y = 7.

m = -coefficient of x/coefficient of y

m = -2/3 ----(1)

We can find slope of the tangent by finding the derivation.

2x + 2y(dy/dx) = 0

2y(dy/dx) = -2x

dy/dx = -x/y ----(2)

(1) = (2)

-2/3 = -x/y

2y = 3x

y = 3x/2

By applying the value of y in the equation of circle, we get

x² + y² = 13

x² + (3x/2)² = 13

x² + (9x²/4) = 13

13x²/4 = 13

x² = 13(4/13)

x² = 4

x = ± 2

So, the required points are (2, 3) (-2, -3).

Example 2 :

At what points on the curve

x² + y² - 2x - 4y + 1 = 0

the tangent is parallel to

(i) x - axis   (ii) y - axis

Solution :

Since the tangent line drawn at the point is parallel to x - axis, slope of line will be equal to 0.

2x + 2y(dy/dx) - 2 - 4(dy/dx) = 0

2x + 2y(dy/dx) - 2 - 4(dy/dx) = 0

2y(dy/dx) - 4(dy/dx) = -2x + 2

(dy/dx)(2y - 4) = 2(-x + 1)

dy/dx = (-x + 1)/(y - 2)

If dy/dx =0

(-x + 1)/(y - 2) = 0

-x + 1 = 0

x = 1

By substituting the value of x = 1, in the given curve, we may solve for y.

x² + y² - 2x - 4y + 1 = 0

1² + y² - 2 - 4y + 1 = 0

y² - 4y = 0

y(y - 4) = 0 

y = 0 and y = 4

So, the required points are (1, 0) and (1, 4).

(ii) y - axis

Solution :

When the tangent line is parallel to y-axis, the slope will become undefined. So, we may consider dy/dx = ∞ or 1/0 

m (or) (dy/dx) = 1/0

dy/dx = (-x + 1)/(y - 2)

(-x + 1)/(y - 2) = 1/0

Doing cross multiplication, we get

0 = y - 2

y = 2

By substituting the value of y in the given equation, we may solve for x.

x² + 2² - 2x - 4(2) + 1 = 0

x² + 4 - 2x - 8 + 1 = 0

x² - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3 and x = -1

So, the required points are (3, 2) (-1, 2).

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