Question

Find the equation of normal to the curve y = x² + 4x + 1 at the point where x = 3 ?
1. x + 10y + 223 = 0
2. x - 10y - 223 = 0
3. x + 10y - 223 = 0
4. None of these
5. x - 10y + 223 = 0

Answer 1

Correct Answer - Option 3 : x + 10y - 223 = 0

Concept:

Let y = f(x) be the equation of a curve, then slope of the tangent at any point say (x1, y1) is given by: \(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\).

Slope of normal at any point say (x1, y1) is given by: \(\frac{{ - 1}}{{\text{Slope of tangent at point}\left( {{x_1},\;{y_1}} \right)}} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}}\)

Equation of tangent at any point say (x1, y1) is given by: \(y - {y_1} = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Equation of normal at any point say (x1, y1) is given by: \(y - {y_1} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Calculation:

Given: Equation of curve is: y = x2 + 4x + 1 

Here, we have to find the equation of normal to the given curve at the point where x = 3.

By substituting x = 3 in the equation y = x2 + 4x + 1 we get,

⇒ y = 3² + 4 ⋅ 3 + 1 = 22.

So, the point of contact is (3, 22)

As we know that slope of normal at any point say (x1, y1) to a curve is given by: \(\frac{{ - 1}}{{\text{Slope of tangent at point}\;\left( {{x_1},\;{y_1}} \right)}}\)

Slope  of the tangent at any point say (x1, y1) to a curve is given by: \(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\)

\(⇒ \frac{{dy}}{{dx}} = 2x + 4\)

\(⇒ {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {3,\;\;22} \right)}} = 2 \cdot 3 + 4 = 10\)

So, slope of normal to the given curve at point (3, 22) is: \( - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} = - {\frac{{1}}{{10}}}\)

As we know that equation of normal at any point say (x1, y1) is given by: \(y - {y_1} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

\(⇒ y - 22 = \; - \frac{1}{{10}} \cdot \left( {x - 3} \right)\)

⇒ x + 10y - 223 = 0

Hence, the equation of normal to the given curve at the point (3, 22) is: x + 10y - 223 = 0