Equation of parabola, tangent at vertex Two tangents on a parabola are $x-y=0$ and $x+y=0$. If $(2,3)$ is the focus of the parabola, then find the equation of tangent at the vertex.
Thanks.
My thoughts:
Can't figure out anything :(
 A: We are given a pair of perpendicular tangents. The point of intersection of tangents lie on directrix of parabola. Clearly, these tangents intersect at origin. Let the equation of directrix be $y=mx$, then the equation of parabola is 
$$(x-2)^2+(y-3)^2=\frac{(y-mx)^2}{1+m^2}$$
Since the parabola is tangent to $y=x$, we get the following quadratic:
$$(x-2)^2+(x-3)^2=\frac{x^2(m-1)^2}{1+m^2}=x^2\left(1-\frac{2m}{1+m^2}\right)$$
The discriminant of the above quadratic must be zero. With this condition, $m=2/3$ and $m=3/2$. Obviously, the second solution is not possible i.e $m=2/3$.
The equation of directrix is hence $y=\dfrac{2}{3}x$. The equation of axis of parabola is perpendicular to directrix and passes through focus. Its equation is $3x-2y-12=0$. The point of intersection of directrix and axis is: $\left(\dfrac{36}{5},\dfrac{24}{5}\right)$. 
The vertex is midpoint of the above point and focus i.e $\left(\dfrac{23}{5},\dfrac{39}{10}\right)$. Hence, the tangent at vertex is:
$$y-\frac{39}{10}=\frac{2}{3}\left(x-\frac{23}{5}\right) \Rightarrow \boxed{4x-6y+5=0}$$
