Given the equation and , determine the range of value of such that the two graphs intersect at exactly two distinct points.
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Hello, Punch! . . . .
Originally Posted by Soroban Hmmm....Soroban, that should be +y^2 / 16 ; right? I end up with something quite different: x = [3a^2 + 128 +- 2aSQRT(17a^2 + 1072)] / (a^2 + 64). Think it's easier to substitute y^2 = (x^2 - 6x + 5) / 4 from [1] to [2].
Originally Posted by Soroban Hello, Punch! . . I tried another method which was to sub equation [1] into [2] and equating the discriminant, but it didn't work. Here's my working.
Your 272a^4 + 17152a^2 simplifies to 16a^2(17a^2 + 1072), which is same as what I showed you in my previous post.
Originally Posted by Wilmer Your 272a^4 + 17152a^2 simplifies to 16a^2(17a^2 + 1072), which is same as what I showed you in my previous post. okay then, maybe we should wait for soroban's reply
Also, is it possible to show how you could transform into ?
Originally Posted by Punch Also, is it possible to show how you could transform into ? That's not what Soroban did: he transformed the other equation. And made (I believe) a sign error; should be (x-2)^2 / a^2 + y^2 / 16 = 1 Multiply that through by 16a^2 and you get: 16x^2 - 64x + 64 +a^2y^2 - 16a^2 = 0
Originally Posted by Wilmer That's not what Soroban did: he transformed the other equation. And made (I believe) a sign error; should be (x-2)^2 / a^2 + y^2 / 16 = 1 Multiply that through by 16a^2 and you get: 16x^2 - 64x + 64 +a^2y^2 - 16a^2 = 0 Sorry! I see it now
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