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Home » Find the equation of the normal to the curve x^2 + 2y^2 – 4x – 6y + 8 = 0 at the point whose abscissa is 2.
Find the equation of the normal to the curve x^2 + 2y^2 – 4x – 6y + 8 = 0 at the point whose abscissa is 2.
CBSE | Class 12 | Exercise 16.2 | Maths | RD Sharma | Tangents and Normals
Find the equation of the normal to the curve x^2 + 2y^2 – 4x – 6y + 8 = 0 at the point whose abscissa is 2.

RD Sharma Solutions for Class 12 Maths Chapter 16 Tangents and Normals Image 70

Finding

    \[y\]

co – ordinate by substituting

    \[x\]

in the given curve

    \[2{{y}^{2}}-\text{ }6y\text{ }+\text{ }4\text{ }=\text{ }0\]

    \[{{y}^{2}}-\text{ }3y\text{ }+\text{ }2\text{ }=\text{ }0\]

Or,

    \[y\text{ }=\text{ }2\text{ }or\text{ }y\text{ }=\text{ }1\]

    \[m\text{ }\left( tangent \right)\text{ }at\text{ }x\text{ }=\text{ }2\text{ }is\text{ }0\]

Normal is perpendicular to tangent so,

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

    \[m\text{ }\left( normal \right)\text{ }at\text{ }x\text{ }=\text{ }2\text{ }is~1/0\]

 which is undefined

Equation of normal is given by

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

    \[x\text{ }=\text{ }2\]

i

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