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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 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.

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