Tangents and normals

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Question

${If the slope of line through the origin is tangent to the curve y}^{} = x^{3} + x + 16 is m ,then the value of m is$

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Solution

$y^{} = x^{3} + x + 16$
${(\frac{d y}{d x})}_{(x_{1}, y_{1})} = 3 x_{1}^{2} + 1$
$3 x_{1}^{2} + 1 = \frac{y_{1}}{x_{1}}$ $= s l o p e o f t h e l i n e p a s \sin g t h r o u g h (0, 0) a n d (x_{1}, y_{1})$
$\Rightarrow 3 x_{1}^{3} + x_{1} = x_{1}^{3} + x_{1} + 16$
$\Rightarrow 2 x_{1}^{3} = 16$
$\Rightarrow x_{1} = 2$ $\Rightarrow y_{1} = 26$
$\Rightarrow$ $m = 13$

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Tangents and normals

Remember concepts with our Masterclasses.

If the slope of line through the origin is tangent to the curve y =x3+x+16 is m ,then the value of m is

y =x3+x+16dydxx1,y1=3x12+13x12+1=y1x1=slope of the line passing through 0,0 and x1,y1⇒3x13+x1=x13+x1+16⇒2x13=16⇒x1=2⇒y1=26 ⇒m=13

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${If the slope of line through the origin is tangent to the curve y}^{} = x^{3} + x + 16 is m ,then the value of m is$