A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

 5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

 

let the rod touch the x axis at A and y axis at B with origin at O

given PA=3

so PB=12-3 = 9

Draw PM perpendicular to the x axis and

PN perpendicular to the y axis

so if P=(x,y) 

PM=y and PN=x

 

Clearly if the rod makes an angle u with the x axis

ie. , if  angle(OAP) = u

then angle(NPB) = u

using triangle APM 

sin u = (y/3)

using triangle BPN

cos u = (x/9)

 

We know that 

[cos u] ^2  + [sin u]^2  = 1

so

(x/9)^2 + (y/3)^2 = 1

[(x^2) / 81]  + [(y^2) / 9] =1

 

Find the area of the triangle formed by the lines joining the vertex of the parabola
x^2 = 12y to the ends of its latus rectum. 

use x^2 = 4ay with x^2 = 12y

to get 4a = 12

or a=12/4 

a=3 

Focus is (0,a)  = (0,3)

so latus rectum = 4a =4*3 = 12

semi latus rectum = 12 / 2 = 6

 

end points of the latus rectum are (6,3)  and (-6,3)

vertex is (0,0)

Now we can either use the formula for area using these three coordinates

given by

 area of a  triangle =

{1/2}[x1(y2-y3) + x2(y3-y1) +x3(y1-y2)]

 

 

OR

 

use the fact that

vertex V=(0,0) ,focus F=(0,3)

height is along the axis VF = 3units

base is along the latus rectum of length =12

so area = (1/2) *base *height

=(1/2)*12*3 =18 square units.

 

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ncert cbse 11th mathematics chapter 11 conic section 

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 5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

solution

Find the area of the triangle formed by the lines joining the vertex of the parabola
x^2 = 12y to the ends of its latus rectum.  

solution 

 

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