What is the curve that always has a normal line with a y-intercept of 6?

[Chris L T521]

Here's this week's problem!

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Problem: Recall that the normal line to a curve at a point $P$ on the curve is the line that passes through $P$ and is perpendicular to the tangent line at $P$. Find the curve that passes through the point $(3,2)$ and has the property that if the normal line is drawn at any point on the curve, then the $y$-intercept of the normal line is always 6.

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[Chris L T521]

This week's problem was correctly answered by chisigma, MarkFL, Opalg, and Pranav. You can find Opalg's solution below.

[sp]If $y'$ denotes the gradient of the curve at the point $(x_0,y_0)$, then the gradient of the normal at that point is $-1/y'$, and the equation of the normal is $y-y_0 = -\dfrac1{y'}(x-x_0), $ or $(y-y_0)y' = -( x-x_0).$ If that line passes through the point $(0,6)$ then $(6-y_0)y' = x_0 .$ If we now drop the subscripts from $x_0$ and $y_0$, then the condition on the curve becomes $(6-y)y' = x .$ That differential equation can be integrated to get $$\int(6-y)\,dy = \int x\,dx$$, or $6y - \frac12y^2 = \frac12x^2 + {}$const. This can be written as $x^2 + (y-6)^2 = {}$const. The condition that the curve passes through $(2,3)$ gives the constant as $25$, so the curve is $x^2 + (y-6)^2 = 25.$

Geometrically, it is obvious that this answer is correct, because the curve is a circle centred at the point $(0,6)$, and any normal to a circle must pass through the centre.[/sp]