## Tangent and Normal Question7

In this page tangent and normal question7 we are going to see solution of some practice questions from the worksheet.

(7) Let P be a point on the curve y = x³ and suppose that the tangent line at P intersects the curve again at Q. Prove that the slope at Q is four times the slope at P.

Solution:

The tangent drawn at the point "P" on the curve touches the curve again at the point "Q"

Let P (a,a³) be a point on the curve

y = x³  --- (1)

dy/dx = 3 x²

slope at the point "P" (dy/dx) = 3 (a)²

= 3 a²

Equation of the tangent at the point "P"

(y - y₁) = m (x - x₁)

(y - a³) = 3 a² (x - a)

y - a³ = 3 a² x - 3

now we are going to apply the value of y in this equation

x³ - a³ = 3 a² x - 3

x³ - a³ - 3 a² x + 3 a³ = 0

x³ - 3 a² x + 2 a³ = 0

(x - a)² (x + 2a) = 0

(x - a)² = 0                       (x + 2a) = 0

(x - a) = 0                                 x = - 2 a

x = a

Here a is the x-coordinate value of the point "P". So we have to take -2a is the x-coordinate value of the point "Q"

Slope at the point Q

dy/dx = 3 x²

= 3 (-2a)²

= 3 (4a²)

= 4 (3a²)

Slope at the point Q = 4 (Slope at the point P)

(8) Prove that the curve 2x²+4y²=1 and 6x²-12y²=1 cut each other at right angles.

Solution:

Let (x₁,y₁) be the common point on the curve

2 x₁² + 4 y₁² = 1     ----- (1)

6 x₁² - 12 y₁² = 1     ----- (2)

to find the point of intersection we have to solve the given equations

(1) x 3 =>    6 x₁² + 12 y₁² = 3

6 x₁² - 12 y₁² = 1

------------------

12 x₁² = 4

x₁² = 4/12

x₁² = 1/3

now we are going to apply the value of x₁² in the first equation

2 (1/3) + 4 y₁² = 1

(2/3) + 4 y₁² = 1

4 y₁² = 1 - 2/3

4 y₁² = 1/3

y₁² = 1/12

Therefore the common point on both curves is (1/3 , 1/12)

differentiate the first equation with respect to x

2 x² + 4 y² = 1

4 x + 8 y(dy/dx) = 0

8 y(dy/dx) = - 4 x

(dy/dx) = - 4 x/8y

= - x/2y

6 x₁² - 12 y₁² = 1

12 x - 24 y (dy/dx) = 0

- 24 y (dy/dx) = -12 x

dy/dx = -12 x/-24y

= x/2y

If two curves are intersecting orthogonally then m₁ x m₂ = -1

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

- x²/4y² = -1

-(1/3)/4(1/12) = -1

- (1/3)/(1/3) = - 1

- 1 = -1

tangent and normal question7 tangent and normal question7