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Question: A point moves such that its distance from the point (4, 0) is half of the distance from the line x =...

A point moves such that its distance from the point (4, 0) is half of the distance from the line x = 16. What is the locus of this point?
(a). 3x2+4y2=1923{x^2} + 4{y^2} = 192
(b). 4x2+3y2=1924{x^2} + 3{y^2} = 192
(c). x2+y2=192{x^2} + {y^2} = 192
(d). None of these

Explanation

Solution

Hint: The formula for distance between two points is (x2−x1)2+(y2−y1)2\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} and distance of a point from a line is ∣ax0+by0+c∣a2+b2\dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}. Use these formulae to equate the given condition and find the locus of the points.

Complete step-by-step answer:
We need to find the locus of all points such that its distance from the point (4, 0) is half of the distance from the line x = 16.
Let this point be (x, y).

The formula for distance between two points (x1,y1)({x_1},{y_1}) and (x2,y2)({x_2},{y_2}) is given as follows:
D=(x2−x1)2+(y2−y1)2D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}
The distance between the points (x, y) and the point (4, 0) is given as follows:
D=(4−x)2+(0−y)2D = \sqrt {{{(4 - x)}^2} + {{(0 - y)}^2}}
Simplifying, we have:
D=(4−x)2+y2..............(1)D = \sqrt {{{(4 - x)}^2} + {y^2}} ..............(1)
The distance of a point (x0,y0)({x_0},{y_0}) from the line ax+by+c=0ax + by + c = 0 is given as follows:
d=∣ax0+by0+c∣a2+b2d = \dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}
Let us write the equation x = 16 in its standard form to apply the above equation.
x+(0)y−16=0x + (0)y - 16 = 0
Then, the distance of the point (x, y) from the line x = 16 is given as follows:
d=∣1.x+0.y−16∣12+02d = \dfrac{{|1.x + 0.y - 16|}}{{\sqrt {{1^2} + {0^2}} }}
Simplifying, we have:
d=∣x−16∣1d = \dfrac{{|x - 16|}}{1}
d=∣x−16∣.............(2)d = |x - 16|.............(2)

Now, the distance of the point (x, y) from the point (4, 0) is half of the distance between the point and the line x = 16. Hence, we have:
D=12dD = \dfrac{1}{2}d
From equation (1) and equation (2), we have:
(4−x)2+y2=12∣x−16∣\sqrt {{{(4 - x)}^2} + {y^2}} = \dfrac{1}{2}|x - 16|
Squaring both sides, we have:
(4−x)2+y2=14(x−16)2{(4 - x)^2} + {y^2} = \dfrac{1}{4}{(x - 16)^2}
Evaluating the squares, we have:
x2−8x+16+y2=14(x2−32x+256){x^2} - 8x + 16 + {y^2} = \dfrac{1}{4}({x^2} - 32x + 256)
Cross-multiplying, we have:
4(x2−8x+16+y2)=x2−32x+2564({x^2} - 8x + 16 + {y^2}) = {x^2} - 32x + 256
Simplifying, we have:
4x2−32x+64+4y2=x2−32x+2564{x^2} - 32x + 64 + 4{y^2} = {x^2} - 32x + 256
Canceling common terms, we have:
4x2+64+4y2=x2+2564{x^2} + 64 + 4{y^2} = {x^2} + 256
Simplifying, we have:
4x2−x2+4y2=256−644{x^2} - {x^2} + 4{y^2} = 256 - 64
3x2+4y2=1923{x^2} + 4{y^2} = 192
Hence, the correct answer is option (a).

Note: You can also find the distance of a point from the line x = 16 without using the formula. It is a vertical line and the distance of any point from this line is the horizontal distance which is ∣x−16∣|x - 16|.