At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point `( 4, 3)`. F

Question

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point `( 4, 3)`. F

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point `( 4, 3)`. Find the equation of the curve given that it passes through `(2, 1)`.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Slope of tangent `=(dy)/(dx)`
Slope of straight line joining the touch point `(x,y)` and the point `(-4,-3)=(y+3)/(x+4)`
Given, `(dy)/(dx)=(2(y+3))/(x+4)`
`implies (dy)/(y+3)=(2dx)/(x+4)`
`implies int(1)/(y+3)dy=2int(1)/(x+4)dx+c`
`implies log(y+3)=2log(x+4)+c`
It passes through the point `(-2,1)`.
`log4=2log2+c`
`implies c=0`
Therefore, `log(y+3)=2log(x+4)`
`implies y+3=(x+4)^(2)`
`implies y=x^(2)+8x+13`