find-the-equation-of-the-2D-curve-such-that-the-lines-x-t-y-a-t-1-are-always-tangent-to-the-curve-given-a-is-a-positive-real-constant-and-t-is-a-parameter-0-lt-t-lt-a-

Question Number 34326 by 33 last updated on 04/May/18

f i n d t h e e q u a t i o n o f t h e 2 D

c u r v e s u c h t h a t t h e l i n e s

\frac{x}{t} + \frac{y}{(a - t)} = 1

a r e a l w a y s t a n g e n t t o

t h e c u r v e .

g i v e n^{'} a^{'} i s a p o s i t i v e r e a l

c o n s t a n t a n d^{'} t^{'} i s a

p a r a m e t e r . (0 < t < a)

Answered by MJS last updated on 05/May/18

we take two of the tangents with

t_{1} = p - h

t_{2} = p + h

and search their intersection

y_{1} = (a - p + h) (- \frac{1}{p - h} x + 1)

y_{2} = (a - p - h) (- \frac{1}{p + h} x + 1)

y_{1} = y_{2} \Rightarrow y_{1} - y_{2} = 0

(a - p + h) (- \frac{1}{p - h} x + 1) - (a - p - h) (- \frac{1}{p + h} x + 1) = 0

2 h (- \frac{a}{p^{2} - h^{2}} x + 1) = 0

x = \frac{p^{2} - h^{2}}{a}

now remember :

first idea to find the tangent in (\begin{matrix} x \\ f (x) \end{matrix}) was

to draw the line between (\begin{matrix} x - h \\ f (x - h) \end{matrix}) and

(\begin{matrix} x + h \\ f (x + h) \end{matrix}) . this is a secant . with h \to 0 we

get the tangent .

same here, just from the opposite of it :

h \to 0 \Rightarrow x = \frac{p^{2}}{a}; y = \frac{{(a - p)}^{2}}{a}

p = \pm \sqrt{a x} \Rightarrow y = x \pm 2 \sqrt{a x} + a

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