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Patel Conics question. (1 Viewer)

dasicmankev

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This is question 20, Exercise 6C.

"Find the equations of the four tangents common to the hyperbola x^2 - 2y^2 = 4 and the circle x^2 + y^2 = 1. Find the points of contact of these tangents with the circle." [Hint: Let xx1 + yy1 = 1 be tangent to x^2 + y^2 = 1 at P(x1, y1)]

Here's what I tried: Equation of tangent to circle: xx1 + yy1 = 1 ----(1)
Equation of tangent to ellipese: xx1/4 - yy1/2 = 1----(2)

yy1 = -1 => y1= -1/y
xx1 = 2 => x1 = 1/2

I then subbed them back into x^2 + y^2 = 1 to obtain 4/x^2 + 1/y^2 = 1
Also subbed it into x^2 - 2y^2 = 4 to get 4/x^2 - 2/y^2 = 4.

Solving these two, I got x= sqrt. 2 and y is a square root of a negative number??? got a bit confused here, someone help me?
 

Affinity

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This is question 20, Exercise 6C.

"Find the equations of the four tangents common to the hyperbola x^2 - 2y^2 = 4 and the circle x^2 + y^2 = 1. Find the points of contact of these tangents with the circle." [Hint: Let xx1 + yy1 = 1 be tangent to x^2 + y^2 = 1 at P(x1, y1)]

Here's what I tried: Equation of tangent to circle: xx1 + yy1 = 1 ----(1)
Equation of tangent to ellipese: xx1/4 - yy1/2 = 1----(2)

yy1 = -1 => y1= -1/y
xx1 = 2 => x1 = 1/2

I then subbed them back into x^2 + y^2 = 1 to obtain 4/x^2 + 1/y^2 = 1
Also subbed it into x^2 - 2y^2 = 4 to get 4/x^2 - 2/y^2 = 4.

Solving these two, I got x= sqrt. 2 and y is a square root of a negative number??? got a bit confused here, someone help me?
(2)'s wrong.. in (1) you assumed that (x1,y1) is on the circle, so you can't use it as a point on the hyperbola.
you should solve (1) and x^2 - 2y^2 = 4 simultaneously and set the discriminant to 0.
Actually.. the better way:

Here's what I tried: Equation of tangent to circle: x[x1] + y[y1] = 1 ----(1)
Equation of tangent to ellipese: x[x2] - 2y[y2] = 4----(2)

now, for them to represent the same line, we must have
[x2]/[x1] = -2[y2]/[y1] = 4
[x2] = 4[x1]
[y2] = -2[y1]
you also know that
[x1]^2 + [y1]^2 = 1
[x2]^2 - 2[y2]^2 = 4

subsitute and simplify
 
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dasicmankev

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Thank you Affinity, that really helped. Yeah, stupid of me not seeing that I had to use (x1, y1) and (x2, y2)...
 

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