MHB Equation of Normal to Curve at (1,5): Solved?

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Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
 
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alexandravo said:
Really confused bout a question and finding the equation.

A normal is drawn at the point (1,5) on the curve defined by the rule y=x2+4. Find the equation of the normal.

I substituted the values x=1 and y=5 into the derived equation and got my answer to be x+2y=10? Is that correct?
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
 
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
 
Country Boy said:
If y=x^2+ 4, y'= 2x. (1, 5) is on the curve so the tangent line through (1, 5) has slope 2(1)= 2.
The tangent line is y(x)= 2x+ C where C is such that y(1)= 2(1)+ C= 5. C= 5- 2= 3. The tangent line is y= 2x+ 3.

You say "I substituted the values x=1 and y=5 into the derived equation". I assume you mean the equation y'= 2x. Yes, with x= 1, y', the slope, is 2 but there is no reason to calculate 2 times 5! y is 5, not x! And y' is the slope of the tangent line, not the constant term.
But we're looking for the normal, not the tangent!
 
Opalg said:
You got the slope right. But the point $(1,5)$ does not lie on the line $x+2y = 10$, so you need to adjust the constant $10$.
Country Boy Is correct in stating that the slope of the tangent line is 2. So the tangent line equation is
y = 2x + 3
and the equation of the normal is then
y = -x/2 + 11/2
 
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