hyperbola.....................?

Question

can anyone pls tell me what is the formula for common tangent,,??.....explain by following sum
1.A common tangent to 9x^2-16y^2=144 and x^2+y^2=9 is...?

2.the equation of tngt to hyperbola 2x^@-3y^2=6 at(3,2) is....?

that is 2x^2....not 2x^@

Answer 1

When one wants a tangent, one usually wants a derivative, and that is the case here. The derivatives of the two expressions give us separate formulas for tangents, and if the tangent is to be common to both, the formulas must be equatable for suitable values of x and y. Try this hint and see how it works out.

Answer 2

Let's begin with the simpler 2nd question.

2. The equation of the tangent to hyperbola
x^2/a^2 - y^2/b^2 = 1 /a = const > 0, b = const > 0/
at point (x0, y0) is the following:
x*x0/a^2 - y*y0/b^2 = 1
Take the hyperbola's equation 2x^2 - 3y^2 = 6 in the form
2x^2/6 - 3y^2/6 = 1, i.e. x^2/3 - y^2/2 = 1, you'll obtain
3x/3 - 2x/2 = 1, i.e. x - y = 1. You can easily convince yourself (solving both equations together) that both lines have indeed only one common point - (3, 2),

1. Proceeding as above, You'll obtain the hyperbola's equation in the form
x^2/16 - y^2/9 = 1, or x^2/4^2 - y^2/3^2 = 1
and a circle x^2 + y^2 = 9, centered at the origin, both situated like ") O (". Now You can guess that there exist 4 common tangents to both curves, encompassing the circle, forming a rhombus like the diamond sign on the playing cards /2 by 2 intersecting on the coordinate axes/. According 2. above, each tangent to the hyperbola may be presented with the equation
x*x0/16 - y*y0/9 = 1
and the question is now: for what x0 and y0 the line with the above equation touches the given circle. By somewhat boring calculations I'll not reproduce here, we obtain
y = ±sqrt(18/7)x ±15/sqrt(7), or
y = 3(±x√2 ± 5)/√7
Taking each of 4 possible combinations of signs (+ and +, + and -, - and +, - and -) we have the equations of 4 tangents. The rhombus, whose sides are the tangents, has vertexes with approximate coordinates (±3.54; 0) and (0; ±5.66).

Answer 3

I do not think there is a formula in text books for the common tangent to the two curves because the formula, you want, does not have common applications.

Whenever one wants the equation to the common tangent can be derived as an exercise for practice.

Answer 4

I would suggest this method
take y =mx+n with m zand n unknown
Intercept with both
example x^2+(mx+n)^2-9=0
(1+m^2)x^2 +2mn x+n^2-9=0
This equation must have a double root

m^2n^2-(1+m^2)(n^2-9)=0 (a)
do the same with the other curve
You get 2 equations for m and n
The problem in general is of fourth degree as normally there
are four common tangents(symmetry may simplify)
Exam. two circles outside each other have four common tangents two externals and two internals

Answer 5

2. Find the equation of the tangent line to the hyperbola
2x² - 3y² = 6 at the point P(3,2).

Differentiate implicitly.

4x - 6y(dy/dx) = 0
dy/dx = -4x/(-6y) = 2x/(3y) = (2*3)/(3*2) = 6/6 = 1

The equation of the tangent line thru P is:

y - 2 = 1(x - 3) = x - 3
y = x - 1
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