Correct Answer - Option 1 : x - y + 5 = 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 = x3 - 2x + 7
Here we have to find the equation of tangent to the curve y = x3 - 2x + 7 at the point (1, 6).
As we know that, slope of the tangent at any point say (x1, y1) is given by: \(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\).
\(⇒ \frac{{dy}}{{dx}} = 3{x^2} - 2\)
\(⇒ {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {1,\;6} \right)}} = 3 ⋅ {\left( 1 \right)^2} - 2 = 1\)
As we know that, equation of tangent at any point say (x1, y1) is given by: \(y - {y_1} = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)
Here, x1 = 1, y1 = 6 and \(⇒ {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {1,\;6} \right)}} = 1\)
⇒ y - 6 = 1 ⋅ (x - 1)
⇒ x - y + 5 = 0
Hence, the equation of the required tangent is: x - y + 5 = 0
