Find the values of $t$ if a straight line $y=t$ is the tangent to the locus of point $P$ , $$x^2+y^2+4x-6y-3=0$$ where $t$ is a constant.
This is an exam question from a school. The solution in the answer scheme is substitute the $t$ into the equation of locus, then use the discriminant (one solution) to find the values of $t$.
I assume the locus touches only one point of the tangent line, $y=t$ , so there is only one solution, and the final answer given is $t=-1$ and $t=7$.
but doesn't the number of solution refer to the number of point that intersect with the x-axis?
So when $y=0$, $$x^2+4x-3=0$$ there should be two solutions, instead of one solution only, the values of $t$ should be $t < -1$ and $t > 7$
So which one is correct?
The two horizontal lines tangent to the ellipse are $y=-1$ and $y=7$ as you found. The horizontal line $y=0$, which is the x-axis, cuts the ellipse in two points as you noticed.
The discriminant is $\Delta = -4(t+1)(t-7)$. It is positive when $t$ lies between $-1$ and $7$, as expected. May be you forgot the minus sign?