Question

A fixed point is 80 mm from a fixed straight line. Draw the locus of apoint P moving such a way that its distance from the fixed straight lineis twice its distance from the fixed point. Draw a normal and tangent atany point on the curve. Name the curves.​

Answer 1

I hope this will help you

Answer 2

Answer:

The given curve is an ellipse.

Step-by-step explanation:

Step 1: Finding the curve and writing the equation

Ellipse is locus of a point such that the distance of point from a fixed point (focus) and a fixed line (directrix) is constant and is less than 1. Hence the given curve is an ellipse.

e = Distance from fixed point / Distance from fixed line

e = 1/2

Let the centre of ellipse be at origin with major axis along x axis.

General equation of ellipse : $x^{2} /a^2 + y^2/b^2 = 1$

Coordinate of fixed point ( focus F ) = (±ae,0)

Fixed line (directrix) cuts x-axis at (± a/e , 0) and is parallel to y-axis

Distance between fixed point and fixed line = a/e - ae

2a - a/2 = 80

a = 160/3

$b = a\sqrt{1-e^{2} }$

b = $80\sqrt{3} /3$

Equation of ellipse : $9x^{2} /25600 + 3y^{2} /6400 = 1$

Step 2: Drawing normal and tangent at a point

Choose point (160/3,0) on ellipse

Normal : y = 0

Tangent : x = 160/3

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