Login/Sign Up
Home » Find the equation of the tangent and the normal to the following curves at the indicated points: (i) y = x2 at (0, 0) (ii) y = 2×2 – 3x – 1 at (1, – 2)
Find the equation of the tangent and the normal to the following curves at the indicated points: (i) y = x2 at (0, 0) (ii) y = 2×2 – 3x – 1 at (1, – 2)
CBSE | Class 12 | Exercise 16.2 | Maths | RD Sharma | Tangents and Normals
Find the equation of the tangent and the normal to the following curves at the indicated points: (i) y = x2 at (0, 0) (ii) y = 2×2 – 3x – 1 at (1, – 2)

(i)

Given

    \[y\text{ }=\text{ }{{x}^{2}}~at\text{ }\left( 0,\text{ }0 \right)\]

By differentiating the given curve, we get the slope of the tangent

RD Sharma Solutions for Class 12 Maths Chapter 16 Tangents and Normals Image 52

    \[m\text{ }\left( tangent \right)\text{ }at\text{ }\left( x\text{ }=\text{ }0 \right)\text{ }=\text{ }0\]

Normal is perpendicular to tangent so,

    \[{{m}_{1}}{{m}_{2}}~=\text{ }-\text{ }1\]

RD Sharma Solutions for Class 12 Maths Chapter 16 Tangents and Normals Image 53

We can see that the slope of normal is not defined

Equation of tangent is given by

    \[y\text{ }-\text{ }{{y}_{1}}~=\text{ }m\text{ }\left( tangent \right)\text{ }\left( x\text{ }-\text{ }{{x}_{1}} \right)\]

    \[y\text{ }=\text{ }0\]

Equation of normal is given by

    \[\begin{array}{*{35}{l}} y\text{ }-\text{ }{{y}_{1}}~=\text{ }m\text{ }\left( normal \right)\text{ }\left( x\text{ }-\text{ }{{x}_{1}} \right) \\ ~ \\ \end{array}\]

    \[x\text{ }=\text{ }0\]

 

(ii)

Given

    \[y\text{ }=\text{ }2{{x}^{2}}-\text{ }3x\text{ }-\text{ }1\text{ }at\text{ }\left( 1,\text{ }-\text{ }2 \right)\]

By differentiating the given curve, we get the slope of the tangent

RD Sharma Solutions for Class 12 Maths Chapter 16 Tangents and Normals Image 54

    \[m\text{ }\left( tangent \right)\text{ }at\text{ }\left( 1,\text{ }-\text{ }2 \right)\text{ }=\text{ }1\]

Normal is perpendicular to tangent so,

    \[{{m}_{1}}{{m}_{2}}~=\text{ }-\text{ }1\]

    \[m\text{ }\left( normal \right)\text{ }at\text{ }\left( 1,\text{ }-\text{ }2 \right)\text{ }=\text{ }-\text{ }1\]

Equation of tangent is given by

    \[y\text{ }-\text{ }{{y}_{1}}~=\text{ }m\text{ }\left( tangent \right)\text{ }\left( x\text{ }-\text{ }{{x}_{1}} \right)\]

    \[y\text{ }+\text{ }2\text{ }=\text{ }1\left( x\text{ }-\text{ }1 \right)\]

    \[y\text{ }=\text{ }x\text{ }-\text{ }3\]

Equation of normal is given by

    \[y\text{ }-\text{ }{{y}_{1}}~=\text{ }m\text{ }\left( normal \right)\text{ }\left( x\text{ }-\text{ }{{x}_{1}} \right)\]

    \[y\text{ }+\text{ }2\text{ }=\text{ }-\text{ }1\left( x\text{ }-\text{ }1 \right)\]

    \[y\text{ }+\text{ }x\text{ }+\text{ }1\text{ }=\text{ }0\]

i

More Material