Minimize f (x, y) = x2 + y2 on the hyperbola xy = 5
A curve generated by intersecting a right circular cone with a plane is called a conic section. If a plane perpendicular to axis of the cone then circle is produced. Plane that is not perpendicular to the axis and that intersects only a single nappe ,then the curve produced is either an ellipse or or parabola . The curve produced by a plane intersecting both nappes then hyperbola is produced. The ellipse and hyperbola are also known as central conics.
Terms to remember
- Vertex: It is the extreme point on a conic section.
- Locus: The set of all points whose coordinates satisfy a given equation.
- Focus: It is a fixed point about which rays are reflected from the curve converge.
- Nappe: It is the one half of double cone.
- Directrix: It is a fixed-line used to construct a conic section.
- Eccentricity: It is a parameter that measures how much the conic section deviates from being circular.
Types of Conic section
There are different types of conic sections formed based on how the cone is cut. It can form a circle, an ellipse, a parabola, or a hyperbola. The below-given diagram shows how different conic sections are formed. Letβs take a look at the sections in more detail with proper definitions.
- Parabolas
It is a type of conic section in which the locus of points in that plane are equidistant from both the directrix and the focus.
Eccentricity e = 1.
General equation: Y = 4aX, a > 0
- Ellipse
It is the set of points in such a way that the sum of the distance from any point on the ellipse to the other fixed point (Foci) is constant.
Eccentricity e < 1.
General equation: X2βa2 + Y2/b2 = 1
- Circle
It is defined as the closed shape formed by tracing a point that moves in a plane such that its distance from a given point is constant.
Eccentricity e = 0.
General equation: X2 + Y2 = a2
- Hyperbola
It is the locus of all those points in a plane in such a way that the difference in their distance from two fixed points (foci) in the plane is a constant. It has two foci and two directrix. It has two asymptotes which are straight lines that form an X(cross) that the hyperbola approaches but never touches.
Eccentricity e > 1.
General equation: X2/a2 β Y2/b2 = 1
Minimize f (x, y) = x2 + y2 on the hyperbola xy = 5.
Solution:
f(X, Y) = X2 + Y2 β’ (i)
XY = 5
Y = 5/X β’ (ii)
f(X) = X2+ (5/X)2 β’ (by putting equation1 into equation 2)
f'(X) = 2X + 25(-2/X3) β’ (1st derivative of equation)
Now, 2X + 25(-2/X3) = 0
2X β 50/X3 = 0
2X4 β 50 = 0
X4 = 50/2
X4 = 25
X2 = Β±β25
X2 = Β±5 β’ [as X2 canβt be -ve]
X = Β±β5
Y = 5/ Β±β5 β’ [put the value of X into equation (ii)
Y = Β±(β5 Γ β5)/ β5
Y = Β±β5
(X, Y) = (β5 ,β5) & (-β5, -β5)
Minimum value: X2 + Y2 = (Β±β5 )2+ (Β±β5)2 = 10
Similar Problems
Question 1: Find the position of the point (6, -5) relative to the hyperbola x2/a2 β y2/b2 = 1?
Solution:
We know that point p(x1, y2) lies
Outside, x2/a2 β y2/b2 β 1 < 0
Inside, x2/a2 β y2/b2 β 1 < 0
So, x12/a2 β y21/b2 β 1
= 62/9 β (-5)2/25 β 1
= 36/9 β 25/25 β 1
= 2
Since 2 > 0
Therefore point (6, -5) lies inside the given equation of hyperbola.
Question 2: Find the eccentricity of the hyperbola 4x2 β 9y2 -8x = 32?
Solution:
4x2 β 9y2 β 8x = 32
4(x2 β 2x ) β 9y2 = 32
4(x2 β 2x + 1) β 9y2 = 32 + 4 [adding 4 both sides]
4(x2 β 2x + 1) β 9y2 = 36
[(x -1)2]/9 β [y2]/4 = 1
a2 = 9 , b2 = 4
e = [1 + b2/a2]1/2
e = (1 + 4/9)1/2
e = (13/9)1/2
e = (β13)/3
Therefore, e = (β13)/3
Question 3: If the center, vertex, and locus of a hyperbola be (0, 0),(4, 0), and (6, 0) respectively, then find the equation of the hyperbola?
Solution:
Center (0, 0)
Vertex (4, 0)
Focus (6, 0)
So, a = 4
ae = 6
e = 6/4 =3/2
For hyperbola,
e = β(1+ b2/a2) = 3/2
9/4 = 1 + b2/16
b2/16 = 5/4
b2 = 20
x2/16 β y2/20 = 1
5x2 β 4 y2 = 80.
Therefore, equation will be 5x2 β 4y2 = 80.
Question 4: If the locus of a hyperbola be 8 and its eccentricity is 3/β5. Find the equation of hyperbola?
Solution:
Eccentricity (e) = 3/β5
b2/a2 = 4/5 β’ (1)
Latus rectum = 8
2(b2)/a = 8
b2/a = 4 β’ (2)
Dividing (2) by (1)
(b2/a2 )/(b2/a) = (4/5)/4
b2 = 20
So, a2 = 25 ,b2=20
Equation of hyperbola:
x2/25 β y2/20 = 1
(4x2 β 5y2)/100 = 1
4x2 β 5y2 = 100
Therefore equation will be 4x2β 5y2 = 100
Question 5: Locus of the feet of the perpendiculars drawn from either focus on a variable tangent to the hyperbola 16y2 β 9x2 = 1?
Solution:
16y2 β 9x2 = 1
9x2 β 16y2 = -1
x2/(1/339) β y2/(1/16) = -1
For conjugate hyperbola equation of auxiliary circle is: x2 + y2 = b2
So, b2 = 1/16
Equation will be x2 + y2 = 1/16
Question 6: Find the length of traverse axis of the hyperbola, when the product of the perpendicular distance from any point on the hyperbola x2/a2 β y2/b2 = 1 having eccentricity e = β3 on its asymptotes is equal to 6.
Solution:
Eccentricity (e)2 = (a2/b2 ) + 1
(β3)2 = (a2/b2) + 1
3 = (a2/b2) + 1
(a2/b2) = 2
(a2) = 2b2
Product of perpendicular = (a2b2)/a2 +b2
[2b2 Γ b2]/(2b2 + b2) = 6
[2b2 Γ b2]/(3b2) = 6
2b2 = 18
b2 = 9
b = Β± 3
Therefore Length = 2 Γ 3 = 6