Question

Prove that the tangents to the curve y^2=2x at the point x=1/2 are right angles

Answer 1

Note:

★ First derivative of a curve y = f(x) at a point gives the slope of tangent at that point.

★ If m1 and m2 are the slopes of lines (or tangent line) L1 and L2 , then ;

• L1 and L2 are parallel if m1 = m2

• L1 and L2 are mutually perpendicular if m1•m2 = -1

Solution:

The given curve is y² = 2x -------(1)

Now,

Differentiating both sides of the given equation with respect to x , we get ;

=>dy²/dx = d(2x)/dx

=> 2y•dy/dx = 2

=> dy/dx = 2/2y

=> dy/dx = 1/y -------(2)

Also,

When x = 1/2 , then using eq-(1) ,

We have ;

=> y² = 2x

=> y² = 2•(1/2)

=> y² = 1

=> y = √1

=> y = ±1

Clearly,

Corresponding to x = 1/2 , we have two real values of y . Thus , at x = 1/2 , there will be two tangents. One tangent at the point (1/2 , 1) and another tangent at the point (1/2 , -1) .

Now,

Using eq-(1) , the slope of the tangent at the point (1/2 , 1) will be ;

=> dy/dx = 1/y

=> dy/dx at (1/2 , 1) = 1/1

=> dy/dx at (1/2 , 1) = 1

=> m1 = 1 (let)

Also,

Using eq-(1) , the slope of the tangent at the point (1/2 , -1) will be ;

=> dy/dx = 1/y

=> dy/dx at (1/2 , -1) = 1/-1

=> dy/dx at (1/2 , -1) = -1

=> m2 = -1 (let)

Now,

m1•m2 = 1•(-1) = -1

Since,

The product of the slopes of both the tangents is –1 , thus we can say that both the tangents are mutually perpendicular..

Thus,

The angle between both the tangents is 90° (right angle) .

Hence proved .