# Find the equations of the tangent and normal to the parabola y^{2} = 4ax at the point (at^{2} , 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 (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1}).

The equation of the given parabola is y^{2} = 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 (at^{2}, 2at) is

dy/dx]_{(at2, 2at)} = 2a/2at = 1/t

Hence, the equation of the tangent at (at^{2}, 2at) is

y - 2at = 1/t (x - at^{2})

⇒ ty - 2at^{2} = x - at^{2}

⇒ ty = x + at^{2}

Now, the slope of the normal at (at^{2}, 2at) is

- 1/slope of the tangent at (at^{2}, 2at)

= - 1/(1/t)

= - t

Thus, the equation of the normal at (at ^{2} , 2at ) is

y - 2at = - t (x - at^{2})

⇒ y - 2at = - tx + at^{3}

⇒ y = - tx + 2at + at^{3}

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

## Find the equations of the tangent and normal to the parabola y^{2} = 4ax at the point (at^{2} , 2at)

**Summary:**

The equation of the tangent and normal to the parabola y^{2} = 4ax at the point (at^{2}, 2at) is y = - tx + 2at + at^{3}

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