The equation of tangent at the point $P$, $Q$ and vertex $A$ of a parabola are $3x+4y-7=0, 2x+3y-10=0$, and $x-y=0$ respectively. I'm trying to show:
- Focus is $(4,5)$
- Length of Latus Rectum is $2\sqrt2$
- Axis is $x+y-9=0$
- Vertex is $(\frac{9}{2}$,$\frac{9}{2})$
I encountered this is a Multiple Choice question where all the answers are correct. My approach was the following:
at vertex $A$, the equation of tangent is $x-y=0$, hence equation of normal is $x+y=t$. This is the axis so focus should pass through it. Let focus be $(4,5)$ then the equation of axis is $x+y-9=0$
Upon solving simultaneous equation, we get the vertex as $(\frac{9}{2}$,$\frac{9}{2})$, and the distance from focus to vertex is $\frac{1}{\sqrt2}$ which when multiplied by 4 is equal to latus-rectum $2\sqrt2$.
Though I got the answers correct but my approach is wrong as I need to find the answer by calculation and not by plugging the value and see the which answers are consistent.
I seem to recall that the image of focus on any tangent line lies on the directix of the parabola. Thus if we take focus as $(a,b)$ we can find three points on the directix. The slope will be parallel to $x-y=0$, and what's perpendicular to directrix will be the axis of the parabola.
However I'm not able to find the correct answer this way.