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Given that the line $x + 3y = 5$ is normal to the curve $y=x^2 + 5x + 6$ at a point $C$, i) find the coordinates of the point $C$ ii) find the equation of the tangent to the curve a the point $C$

Darshan
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N.vANS
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    http://meta.math.stackexchange.com/questions/12832/dealing-with-zero-effort-questions – Santropedro Mar 16 '17 at 02:45
  • You must do effort to solve the problem, improve the title so it describes the question. People here will not work for you if you show so little effort. – Santropedro Mar 16 '17 at 02:46
  • What have you done so far? What can you say about point $C$? Obviously, since you need a point, you need two coordinates, that means two equations involving $x$, and $y$ – Andrei Mar 16 '17 at 02:47
  • I see you are new to MSE,no worries,see if you have described some of your approach and your title is relevant enough then why not any potential user will not answer your question.So in order to get fruitful response edit your title to make it relevant and post your question with your approach , how you tried it to solve and where exactly you are facing problems.Good luck! – BAYMAX Mar 16 '17 at 02:54

2 Answers2

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$y=x^2+5x+6$

$\dfrac{dy}{dx}=2x+5$ let's say slope $(m_1)=2x+5$

From line equation slope $($say $m_2$$)$= $\dfrac{-1}{3}$

Now for lines to pe perpendicular $m_1 \times m_2 = -1$

From here can you do?

Noddy
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Part 1) Put $y=\frac{5-x}{3}$ in the equation of curve and solve the quadratic equation for $x$ and then substitute for $y$

Now slope of your normal is $\frac{-5}{3}$, slope of your tangent will be $\frac{3}{5}$

Rayees Ahmad
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