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Question: How do you find the equation of a normal line to a curve at the given point?...

How do you find the equation of a normal line to a curve at the given point?

Explanation

Solution

Here we are given a curve and the point at which normal is to be found. So we can find the derivative of the function at that point. This will be slope of tangent and we know that slope of normal=−1slope of tangent{\text{slope of normal}} = - \dfrac{1}{{{\text{slope of tangent}}}} and therefore we will get the equation of the normal as y=mx+by = mx + b and bb can be found by satisfying the equation with the point.

Complete step by step solution:
Here we are given that we need to find the equation of the normal line to a curve at the given point. This equation of the normal will be of form y=mx+by = mx + b and in this mm is the slope of the normal line to the curve. This can be found by finding the negative reciprocal of the slope of the tangent which can be given as slope of normal=−1slope of tangent{\text{slope of normal}} = - \dfrac{1}{{{\text{slope of tangent}}}} and therefore we will get the slope of the normal and then we just need to find the value of the constant bb.
This bb can be found by substituting the value of mm and the point given through which the curve is passing.
Here this can be made clearer with one example.
Let us suppose that we need to find the equation of the normal line to the curve y=x2y = {x^2} at the point (2,4)\left( {2,4} \right)
So we can differentiate the above function and we will get:
f′(x)=2xf'\left( x \right) = 2x
At the point (2,4)\left( {2,4} \right) we can say that f′((2,4))=2(2)=4f'\left( {\left( {2,4} \right)} \right) = 2\left( 2 \right) = 4
Now this is the slope of the tangent at the given point on the curve. So now we can find the slope of the normal and we can apply slope of normal=−1slope of tangent{\text{slope of normal}} = - \dfrac{1}{{{\text{slope of tangent}}}}
slope of normal(m)=−14{\text{slope of normal}}\left( m \right) = - \dfrac{1}{4}
So we get the equation as:
y=mx+by = mx + b
y=−14x+by = - \dfrac{1}{4}x + b
Now we know that this line of normal is at point (2,4)\left( {2,4} \right) hence it need to satisfy this equation and we will get:
y=−14x+by = - \dfrac{1}{4}x + b
4=−14(2)+b b=4+12=92  4 = - \dfrac{1}{4}\left( 2 \right) + b \\\ b = 4 + \dfrac{1}{2} = \dfrac{9}{2} \\\

Hence we get the equation as y=−14x+92y = - \dfrac{1}{4}x + \dfrac{9}{2}

In this way, we can easily find the equation of the normal line at any point on the given curve.

Note:
Here even if we are given to find the equation of the tangent, we just need to find the derivative of the function and this will be equal to the slope of the tangent and the value of bb can be found in a similar way by substituting the point and value of the slope.