How do you find the equations for the normal line to #x^2/32+y^2/8=1# through (4,2)?

2 Answers
Dec 23, 2016

The normal line to the curve through the point #(4,2)# has equation:

#y=2x-6#

Explanation:

For a curve in implicit form:

#F(x,y) = 0#

the normal line to the point #(barx,bary)# is given by the formula:

#(delF)/(dely)(barx,bary)(x-barx)-(delF)/(delx)(barx,bary)(y-bary) = 0#

We have:

#F(x,y) = x^2/32+y^2/8-1#

#(delF)/(delx) = x/16#

#(delF)/(dely) = y/4#

so that the line normal to the curve in #(4,2)# has equation:

#2/4(x-4)-4/16(y-2)=0#

or multypling everything by 4 to have integral coefficients:

#2(x-4)-(y-2) = 0#

#2x-8-y+2=0#

#y=2x-6#

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Dec 24, 2016

#y=2x-6#

Explanation:

The tangent at #(x',y')# to a conic section is given by a substitution rule, which in this case yields #(x x')/32+(y y')/8=1#, which for #x'=4#, #y'=2# may be re-written as #x+2y=8#. Consequently the normal will be #2x-y=c# (using #m_1 xx m_2=-1#) where #c# is evaluated by substituting #x=4#, #y=2#, giving #8-2=6#. Hence the normal is #2x-y=6#, re-arrangeable to #y=2x-6#.

This works for any conic section anywhere, with suitable further substitution rules. More complex rules give chords.