GCSE; Find the possible value of m

Asked Oct 31 '21 at 12:27

Active Oct 31 '21 at 13:18

Viewed 42 times

Hello everyone! I hope someone could help me with this exercise! I tried to find the solution but I didn’t find the right response.

My atempt was the following:

I have substituted $y=mx + 2$ Group the terms like so:

$(1+m^2) x^2 + (4-2m)x + 2 = 0$

This is now of the form $ax^2 + bx + c = 0$ where $a = 1+m^2$, $b = 4-2m$, $c = 2$

But I don’t know what to do next...

I set the discriminant of the quadratic (in terms of m) to zero, and I solve it! And the right solution was m = -2 + root 6 Thank you all!

edited Oct 31 '21 at 13:18

asked Oct 31 '21 at 12:27

Ko_17

1

Can you show your attempt? It would help a lot in terms of helping you correct your approach – Stephen Donovan Oct 31 '21 at 12:30
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Two curves being tangent to each other means that they intersect at exactly one point. Intersection is the graphical interpretation of solving algebraically. So, figure out the condition for which the two curves have exactly one solution. And please use mathjax to format your post. – ryang Oct 31 '21 at 12:43
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Hint: Try using the discriminant approach. – Dumb Mathematician Oct 31 '21 at 12:46
Thank you for your responses!! – Ko_17 Oct 31 '21 at 13:00
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Does this answer your question? Equation of tangents from external point to a circle Once you have the general case, you can apply it to any problem of this form, even if it does use matrices. Found using Approach0. – Toby Mak Oct 31 '21 at 13:11
$m$ is one single constant. The question is, what should the values of $x$ be. Just solve for $x$ and interpret $\pm$ sign. – Narasimham Nov 02 '21 at 13:48

0 Answers0