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Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 9 Maths Textbook Solution.

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ANSWER: Equation of tangent: y+\left ( \frac{71}{4} \right )=3\left \{ x+\left ( \frac{1}{2} \right ) \right \}


 Differentiate  with respect to x   to get its slope


y=x^{2}+4x-16 Which is parallel to 3x-y+1=0


Upon differentiation

\frac{dy}{dx}=2x+4\Rightarrow m\left ( tangent \right )=2x+4

The equation of tangent is given by y-y_{1}=m\left ( tangent \right )\left [ x-x_{1} \right ]

Comparing the slopes of a tangent with the given equation

2x+y=3\Rightarrow x=-\frac{1}{2}

Substitutes the value of x  in the curve to find y

y=\left ( \frac{1}{4} \right )-2-16=-\frac{71}{4}

So,the equation of tangent is parallel to the given line is

y+\left ( \frac{71}{4} \right )=3\left \{ x+\left ( \frac{1}{2} \right ) \right \}

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