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# Solution for Find the Equation of the Tangent Line to the Curve Y = X2 + 4x − 16 Which is Parallel to the Line 3x − Y + 1 = 0 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Find the equation of the tangent line to the curve y = x2 + 4x − 16 which is parallel to the line 3x − y + 1 = 0 ?

#### Solution

Let (x0y0) be the point of intersection of both the curve and the tangent.

$y = x^2 + 4x - 16$

$\text { Since }, \left( x_0 , y_0 \right) \text { lies on curve . Therefore }$

$y_0 = {x_0}^2 + 4 x_0 - 16 . . . \left( 1 \right)$

$\text { Now,} y = x^2 + 4x - 16$

$\Rightarrow \frac{dy}{dx} = 2x + 4$

$\text { Slope of tangent} = \left( \frac{dy}{dx} \right)_\left( x_0 , y_0 \right) =2 x_0 +4$

$\text { Given that The tangent is parallel to the line So,}$

$\text { Slope of tangent=slope of the given line}$

$2 x_0 + 4 = 3$

$\Rightarrow 2 x_0 = - 1$

$\Rightarrow x_0 = \frac{- 1}{2}$

$\text { From} (1),$

$y_0 = \frac{1}{4} - 2 - 16 = \frac{- 71}{4}$

$\text { Now, slope of tangent},m=3$

$\left( x_0 , y_0 \right) = \left( \frac{- 1}{2}, \frac{- 71}{4} \right)$

$\text { Equation of tangent is }$

$y - y_0 = m \left( x - x_0 \right)$

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

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

$\Rightarrow 4y + 71 = 12x + 6$

$\Rightarrow 12x - 4y - 65 = 0$

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Solution for question: Find the Equation of the Tangent Line to the Curve Y = X2 + 4x − 16 Which is Parallel to the Line 3x − Y + 1 = 0 ? concept: Tangents and Normals. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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