We will learn how to find the position of a point in relation to the hyperbola.

The point P (x (_ {1} ), y (_ {1} ))
lies outside, on or within the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 to ( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 < 0, = oder > 0.

Let P (x (_ {1} ), y (_ {1} )) be an arbitrary point on the plane of the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 ………………… .. (me)

Draw from point P (x (_ {1} ), y (_ {1} )) PM perpendicular to XX ‘(i.e. x-axis) and meet the Hyperbola at Q.

From the diagram above we can see that the points Q and P have the same abscissa. Therefore the coordinates of Q are (x (_ {1} ), y (_ {2} )).

Since the point Q (x (_ {1} ), y (_ {2} )) on the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1.

Therefore,

( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {2} ^ {2}} {b ^ {2}} ) = 1

( frac {y_ {2} ^ {2}} {b ^ {2}} ) = ( frac {x_ {1} ^ {2}} {a ^ {2}} ) – 1 ………………… .. (i)

Point P is now outside, on or within the hyperbole

after how

PM <, = oder > QM

ie after y (_ {1} ) <, = oder > y (_ {2} )

ie after how ( frac {y_ {1} ^ {2}} {b ^ {2}} )<, = oder > ( frac {y_ {2} ^ {2}} {b ^ {2}} )

ie after how ( frac {y_ {1} ^ {2}} {b ^ {2}} )<, = oder > ( frac {x_ {1} ^ {2}} {a ^ {2}} )– 1, [Using (i)]

ie after how ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} )<, = oder > 1

ie after how ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 <, = oder > 0

Hence the point

(I) P (x (_ {1} ), y (_ {1} )) lies outside the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 if PM

ie, ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 <0.

(ii) P (x (_ {1} ), y (_ {1} )) lies on the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 if PM = QM

ie, ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 = 0.

(ii) P (x (_ {1} ), y (_ {1} )) lies within the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 if PM

ie, ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} )
– 1> 0.

The point P (x (_ {1} ), y (_ {1} )) lies outside, on or inside the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} )

= 1 according to x ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} )
– 1 <, = oder > 0.

Note:

Assume E (_ {1} ) = ( frac {x_ {1} ^ {2}} {a ^ {2}} ) ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1, then the point P (x (_ {1} ), y (_ {1} )) lies outside, on or inside the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 according to E (_ {1} ) <, = oder > 0.

Solved examples to determine the position of the point (x (_ {1} ), y (_ {1} )) with respect to a hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1:

1. Determine the position of the point (2, – 3) in relation to the hyperbola ( frac {x ^ {2}} {9} ) – ( frac {y ^ {2}} {25} ) = 1.

Solution:

We know the point (x (_ {1} ), y (_ {1} )) lies outside, on or inside the hyperbola ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 according to

( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 < , = oder > 0.

For the given problem we have

( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 = ( frac {2 ^ {2}} {9} ) – ( frac {(- 3) ^ {2}} {25} ) – 1 = ( frac {4} {9} ) – ( frac {9} {25} ) – 1 = – ( frac {206} {225} ) <0.

Therefore the point (2, – 3) lies outside the hyperbole ( frac {x ^ {2}} {9} ) – ( frac {y ^ {2}} {25} ) = 1.

2. Determine the position of the point (3, – 4) in relation to the hyperbole ( frac {x ^ {2}} {9} ) – ( frac {y ^ {2}} {16} ) = 1.

Solution:

We know the point (x (_ {1} ), y (_ {1} )) lies outside, on or within the hyperbole ( frac {x ^ {2}} {a ^ {2}} ) – ( frac {y ^ {2}} {b ^ {2}} ) = 1 according to

( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 < , = oder > 0.

For the given problem we have

( frac {x_ {1} ^ {2}} {a ^ {2}} ) – ( frac {y_ {1} ^ {2}} {b ^ {2}} ) – 1 = ( frac {3 ^ {2}} {9} ) – ( frac {(- 4) ^ {2}} {16} ) – 1 = ( frac {9} {9} ) – ( frac {16} {16} ) – 1 = 1 – 1 – 1 = -1 <0.

Therefore the point (3, – 4) lies outside the hyperbole ( frac {x ^ {2}} {9} ) – ( frac {y ^ {2}} {16} ) = 1.

The hyperbole

11th and 12th grade math degree

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