Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at² , 2at)

Solution:

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

For a curve y = f(x) containing the point (x₁,y₁) the equation of the tangent line to the curve at (x₁,y₁) is given by 

y − y₁ = f′(x₁) (x − x₁).

The equation of the given parabola is y² = 4ax

On differentiating both sides with respect to x, we have:

2y dy/dx = 4a

⇒ dy/dx = 2a/y

Therefore, the slope of the tangent at (at², 2at) is

dy/dx]_{(at², 2at)} = 2a/2at = 1/t

Hence, the equation of the tangent at (at², 2at) is

y - 2at = 1/t (x - at²)

⇒ ty - 2at² = x - at²

⇒ ty = x + at²

Now, the slope of the normal at (at², 2at) is

- 1/slope of the tangent at (at², 2at)

= - 1/(1/t)

= - t

Thus, the equation of the normal at (at ² , 2at ) is

y - 2at = - t (x - at²)

⇒ y - 2at = - tx + at³

⇒ y = - tx + 2at + at³

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 22

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at² , 2at)

Summary:

The equation of the tangent and normal to the parabola y² = 4ax at the point (at², 2at) is y = - tx + 2at + at³