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.

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

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

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)

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

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

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

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.

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.

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

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

(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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