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Question regarding tangent and curves

How do I find the value of k for which 8y=x+2k is a tangent to the curve 2y^2=x+k.

The answer is 8 but I can't seem to get to it!

Thanks in advance :smile:

Reply 1

Swayum

Two curves are tangent at a point if, when you try solving them together, there's a double root at that point. So substitute 8y = x + 2k into 2y^2 = x + k. Rearrange it a bit. Then work out at which point there'll be a double root. Solve for k.

Reply 2

summersolitude

Swayum

I got k=2y^2-8y. Is that correct? Where to I go from here?

I tried substituting the y and I ended up with 0 which is wrong.

Reply 3

Swayum

summersolitude

I got k=2y^2-8y. Is that correct? Where to I go from here?

Nope.

x = 8y - 2k

So 2y^2 = x + k = 8y - 2k + k = 8y - k

(k = 8y - 2y^2 if you like)

Let's go back to what I said. "when you try solving them together, there's a double root at that point"

We've done the "when you try solving them together" part.

Now when will 2y^2 - 8y + k = 0 have a double root? Hint: it's a quadratic isn't it? When do quadratics have double roots?