Draw the locus of the point p moving so that the ratio of the distance from a fixed point F through its distance from a fixed straight line DD' is 3/4 .Point F is distance of 70mm from DD'. Draw the tangent and normal at any point on the curve?
Answers
Answer:
Step-by-step explanation:
Let ax+by+c=0 be the fixed line.
be a fixed point (h, k)
The fixed point's distance from the line is (ah+bk+c)/(a2+b2).
As a result, (ah+bk+c)/(a2+b2)=+/-60.
Bringing both sides together
(ah+bk+c)^2/(a^2+b^2)=3600…………..(1)
If x', y' is a moving point on a curve, then squaring the distances from a fixed point and a fixed line, and equating with a square of 2:3, i.e. 4:9, is the solution.
(x’-h)^2+(y’-k)^2=4/9*(ax’+by’+c)^2/(a^2+b^2)………..(2)
By multiplying sides and cancelling the common term in equations 1 and 2,
((x’-h)^2+(y’-k)^2)(ah+bk+c)^2=4/9*3600(ax’+by’+c)^2
Because a, b, c, and h k are constants, the only variables are x' and y'.
We get locus by removing dashes from x' and y'.
((x-h)^2+(y-k)^2)(ah+bk+c)^2=1600(ax+by+c)^2
Terms for x2, xy, y2, x, y, and constants can be expanded and rearranged.It's a circle, and tangents/normals can be added as needed.
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It is a conic section with equation
Given that the ratio of distance from a fixed point to distance from a straight line is . the distance from the line to fixed point is 70mm.
To draw the locus of point,
Consider the fixed line to be y axis, then the fixed point is (70,0)
The vertex of the locus lies at (40,0).
As,
Let P(x,y) be any point on the locus then,
Distance from fixed point =
Distance from fixed line= x
is the equation of locus of point P moving so that the ratio of distance from a fixed point to distance from a straight line is 3/4 and distance from fixed point to fixed line is 70mm
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