Question

for what values of 'a' will y=ax be a tangent to x^2+y^2+20x-10y+100=0

Can anyone help me with this question. Thanks.

Answer 1

jeeZ...have any ideas?

Answer 2

Is this an x^3 or an x^2?

Answer 3

sorry about that. it was x^2

Answer 4

substitute y=ax in x^2 + y^2....=0
you will get a quadratic equation in x

Answer 5

since y=ax intersects x^2 + Y^2+...=0
at only one point,
the discriminant should be zero.

Answer 6

in other words, we should have repeated roots for the quadratic equation in x.

Answer 7

Good one. Do you know how to solve the problem with these hints?

Answer 8

You will have something like this,
\[(1+a^2)x^2+(20-10a)x+100=0\]that you can solve with the quadratic equation

Answer 9

You will obtain,
\[x=\frac{5 \left(a-2-\sqrt{-3 a^2-4 a}\right)}{a^2+1}\] And
\[x=\frac{5 \left(a-2+\sqrt{-3 a^2-4 a}\right)}{a^2+1}\] But there must be only one point so,
\[\sqrt{-3 a^2-4 a}=0\Rightarrow a=-4/3 ; a=0\]

Answer 10

i graphed it, the answer is correct..
looks something like this