Question

Find a curve y=y(x) through (1, 2) such that the tangent to the curve at any point (xo, y(xo)) intersects the x axis at xI=xo/2.

Answer 1

@Luigi0210 ?

Answer 2

Sorry, don't know @satellite73 @ganeshie8 @Directrix

Answer 3

eq. of any line through ( x0,y(x0)) is
y-y(x0)=m(x-x0), where m is the slope.
because it passes through\[(\frac{ xo }{ 2 },0)\]
\[0-y(x0)=m(\frac{ x0 }{ 2 }-x0)\]
\[-y(x _{0})=-m \frac{ x _{0} }{ 2 }\]
\[\frac{ 2y \left( x _{0} \right) }{ x _{0} }=m ~at ~(x _{0},y(x _{0}))\]
\[m=\frac{ dy }{ dx }=\frac{ 2y }{ x }\]
separating the variables and integrating
\[\int\limits \frac{ dy }{ y }=2\int\limits \frac{ dx }{ x }+C\]
\[ \ln y=2 \ln x+C\]
\[\ln y=\ln x^2+C\]
\[\ln y-\ln x^2=C\]
\[\ln \frac{ y }{ x^2 }=C\]
\[\frac{ y }{ x^2 }=e^C=c (say)\]
\(because~ it~ passes ~through~ (1,2)\)
\[\frac{ 2 }{ 1^2 }=c,c=2\]
?

Answer 4

Thank you!

Answer 5

yw