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Examples on Finding Locus of Point

Example:

C is a centre of the hyperbola x2/a2–y2/b2 = 1 and the tangent at any point P meets asymptotes in the point Q and R. Find the equation to locus of the centre of the circle circumscribing the triangle CQR.
This equation to the hyperbola is given as

 

         x2/a2–y2/b2 = 1                                      …… (1)

 

Let P any point on it as (a sec ?, b tan ?), then the equation of tangent at P is

 

        x/a – y/b sin ? = cos ?                              …… (2)

 

 

The equation to the asymptotes to (1) are

 

        x/a = y/b                                                …… (3)

 

and   x/a = – y/b                                             …… (4)

Solving (2) and (3), we get the coordinates of Q as

        (acos?/1–sin?, bcos?/1–sin?)

 

Solving (2) and (4), we get the coordinates of R as

 

        (acos?/1+sin?, –bcos?/1+sin?)

 

Let O be the centre of the circle passing through C, Q and R having its coordinates as (h, k). Then clearly OC = OQ

 

⇒ h2 + k2 = (h–acos?/1–sin?)2 + (k–bcos?/1–sin?)2

 

⇒ h2 + k2 = (h–acos?/1–sin?)2 + (k–bcos?/1–sin?)2

 

⇒ h2 + k2 = h2 + k2 + (a2 + b2) cos2?/(1–sin?)2 – (2ah + 2bk) cos?/1–sin?

 

⇒ 2(ah + bk) = (a2 + b2) cos?/1–sin?                              …… (5)

 

Similarly OC = OR

 

Hence h2 + k2 = (h–acos?/1+sin?)2 + (k+bcos?/1+sin?)2

 

Which on simplification as in the last case, given

 

        2(ah –bk) = (a2 + b2) cos?/1–sin?                                   …… (6)

 

to get the locus of the point O we have to eliminate f from (5) and (6), so multiplying the two we get

 

       4(a2h2 – b2k2) = (a2 + b2) cos2?/1–sin2? = (a2 + b2)2
for (h, k), we get the required locus as

 

        4(a2x2 – b2y2) = (a2 + b2)2

 

Example:

 

A straight line is drawn parallel to the conjugate axis of a hyperbola meets it and the conjugate hyperbola in the points P and Q. Find the locus of point of intersection of tangents at P and Q.

 

Solution:

 

Let the equation to the hyperbola be

 

                x2/a2–y2/b2 = 1                                      …… (1)

 

and its conjugate hyperbola be

 

                y2/b2–x2/a2 = 1                                      …… (2)
Let p be any point (a sec ?, b tan ?) on P. The equation of the line parallel to the conjugate axis (1) i.e. y-axis passes through P will be 
x = a sec ?                                                                 …… (3)

 

The line (3) will cut the conjugate hyperbola (2) at Q where x = a sec ? and hence y = b √(1+sec2?), therefore the coordinates of Q will be {asec?, b√(1+sec2?)}

 

Now the equation to the tangent to (1) at P is

 

        x/a–y/b sin ? = cos ?

 

or     x/a – cos ? = y/b sin ?                                      …… (4)

 

and the equation to the tangent to (2) at Q is

 

        y/b √(1+sec2?) sec ? = 1

 

⇒ x/a + cos ? = y/b √(1+cos2?)                                  …… (5)

 

 

on squaring and adding (4) and (5), we have

 

         2x2/a2 + 2 cos2 ? = y2/b2 [(1 + cos2 ?) + sin2 ?] = 2 y2/b2
⇒ cos2 ? = y2/b2 – x2/a2

 

Putting the value of cos ? in (5) we get

 

         x/a + √(y2/b2–x2/a2) = y/b √(1+y2/b2–x2/a2)
Squaring we have,

 

         x2/a2 + y2/b2 – x2/a2 + 2 x/a √(y2/b2 – x2/a2) = y2/b2 (1+y2/b2–x2/a2)

 

⇒ 2x/a √(y2/b2–x2/a2) = y2/b2(y2/b2–x2/a2)

 

⇒ y2/b2 √(y2/b2–x2/a2) = 2x/a

 

⇒ y4/b4 (y2/b2–x2/a2) = 4x2/a2

 

There is the required locus.
Example:

From a point A, perpendiculars AB and AC are drawn to two straight lines OB and OC. If the area OBAC is constant, find the locus of A.

Solution:

Let the bisectors of the angles BOC be taken as axis. So the equations of OB and OC are respectively.

         x cos α + y sin α = 0

 

and   x cos α – y sin α = 0            where α = 1/2 ∠BOC

 

Take any point A as (h, k); then

 

AB = Perpendicular from A on OB
        = hcosα+ksinα/√cos2α+sin2α = h cos α + k sin α      …… (1)

 

 

and similarly

 

AC = Perpendicular from A on C

 

        = h cos α – k sin α                                              …… (2)

 

The equation to AB will be

 

        (h – x) sin α + (y – k) cos α = 0
⇒ y cosα – x sinα + h sin α – k cos α = 0                       …… (3)
Similarly the equation AC will be

 

 

        (h – x) sin α – (y – k) cos α = 0

1698_Perpendicular distance.JPG

 Now OB = Perpendicular distance of (3) from (0, 0).

 

 

        = 0–0+hcosα+ksinα/√cos2α+sin2α

 

        = h sin α – k cos α

 

Similarly OC = perpendicular distance of (0, 0) from (4)

 

        = h sin α + k cos α.

 

Now the area of quad. OBAC = ΔOAB + ΔOAC

 

        = 1/2 OB × AB + 1/2 OC × AC

 

        = 1/2 [h sinα – kcosα][h cosα + k sinα] + 1/2 [h cosα – k sinα] [h sinα + k cosα]

 

        = (h2 – k2) sinα.cosα = constant = S (say)

 

⇒ h2 – k2 = {s/sina, cosa} which is again constant = a2 (say)

Therefore the locus of the point (h, k) will be x2 – y2 = a2, which is hyperbola.

To read more, Buy study materials of Hyperbola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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