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202211041452 Solution to 1972-CE-AMATH-I-10

A point P moves such that its distance from the point (1,5) is equal to its distance from the line y=x. Write down and simplify the equation of the locus of P.

Roughwork.

Let the locus of P be t-parametrized as (x_P(t),y_P(t))=(t,f(t)), and let the line L:y=x be s-parametrized as (x_L(s),y_L(s))=(s,s).

For the shortest line joining any point P on the locus to its perpendicular projection on L, its slope is:

\textrm{Slope}_\alpha = \displaystyle{\frac{-1}{y_L'(s)} = \frac{-1}{1} = -1}

with angle of inclination \displaystyle{\alpha =\frac{3\pi}{4}}, and for the line joining any point P on the locus to point (1,5), its slope is

\textrm{Slope}_\beta =\displaystyle{\frac{y_P(t_0)-5}{x_P(t_0)-1}},

with angle of inclination \displaystyle{\beta =\tan^{-1}\bigg(\frac{y_P-5}{x_P-1}\bigg)}, whereas the slope of tangent at any point (x_P(t_0),y_P(t_0)) on the locus is

\textrm{Slope}_\gamma =f'(t_0)=\displaystyle{\frac{\mathrm{d}f(t)}{\mathrm{d}t}\bigg|_{t=t_0}}.

with angle of inclination \gamma =\tan^{-1}(f'(t_0)).

Have a picture in mind

Recall

\displaystyle{\tan (A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}}.

Hence,

\begin{aligned} \tan (\gamma -\alpha ) & = \tan (\gamma -\beta ) \\ \frac{\tan\gamma -\tan\alpha}{1+\tan\gamma\tan\alpha} & = \frac{\tan\gamma -\tan\beta}{1+\tan\gamma\tan\beta} \\ \frac{(f'(t_0))-(-1)}{1+(f'(t_0))(-1)} & = \frac{(f'(t_0))-\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)}{1+(f'(t_0))\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)} \\ \end{aligned}

The remaining are left the reader as an exercise.