0
$\begingroup$

Find the y-coordinate of all points on the curve $2x + (y+2)^2=0$ where the normal line to the curve passes through the point (-27,-50) (not on curve).

There are 3 answers.

I started by taking the derivative of the function and got: $$dy/dx=mtan=-1/(y+2)$$

So the slope of the normal is $y+2$

I then used the point in the slope=slope formula and got the following equation for the normal line $$xy+26y+2x+4$$ Which I then set equal to the original equation to find the points of intersection, and ended up with $$y^2-xy-22y$$ The general formula for the y values is x + 22, and I don't know how to get 3 values from this.

$\endgroup$
  • $\begingroup$ Call the point of "normalcy" $(a,b)$, not $(x,y)$. $\endgroup$ – André Nicolas Oct 26 '13 at 23:11
1
$\begingroup$

Hint: The line connecting $\left(-\dfrac{(y+2)^2}{2}, y\right)$ and $(-27,-50)$ has slope $y+2$: $$\frac{-(y+2)^2/2+27}{y+50} = y+2 \quad \iff \quad -(y+2)^2 + 54 = 2(y+2)(y+50).$$ This gives you two solutions. For the third one note that slope of the normal line may also be equal to $\infty$.

$\endgroup$
0
$\begingroup$

You have the correct expression for the slope of the normal line. You then need to state that at the point $(x_1, y_1)$ on the original curve, the slope of the normal line will be given by:$$m=y_1+2\tag{1}$$and the point $(x_1,y_1)$ will satisfy the original equation, i.e.:$$2x_1+(y_1+2)^2=0$$This leads to:$$x_1=-\frac{(y_1+2)^2}{2}\tag{2}$$So the equation of the normal line will be given by:$$y-y_1=m(x-x_1)$$Now use equations (1) and (2) to substitute $m$ and $x_1$, then use the fact that the line passes through $(-27,-50)$ to solve for $y_1$. That should give you your 3 values.

$\endgroup$
0
$\begingroup$

There is a similar problem here, so I have submitted a solution for this one at that location (the three points aren't as awful to solve for as the situation appears)...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.