Math, asked by januaryjockerkng, 6 months ago

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

Answered by syed2020ashaels
0

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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Answered by Rameshjangid
0

It is a conic section with equation 7x^{2} +16y^{2}-2240x+78400=0

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 =\sqrt{(x-70)^{2}+y^{2}

Distance from fixed line= x

\frac{distance .from. fixed .point}{distance .from.fixed. line} =\frac{3}{4}\\\frac{\sqrt{(x-70)^{2}+y^{2}  } }{x} =\frac{3}{4} \\\frac{(x-70)^{2}+y^{2}  }{x^{2} }=\frac{9}{16}\\ 16(x-70)^{2}+16y^{2}  =9x^{2} \\16(x^{2} -2*70x+70^{2} )+16y^{2} =9x^{2} \\16x^{2} -2240x+78400+16y^{2}=9x^{2} \\7x^{2} +16y^{2}-2240x+78400=0

7x^{2} +16y^{2}-2240x+78400=0 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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