If the equation of tangent at point $P(3,4)$ on the parabola whose axis is x-axis is $3x-4y+7=0 $ then find the distance of the tangent from the focus of the parabola.
Method $1$:
I let the equation of parabola to be $y^2=4a(x-h)$
Putting $(3,4),$ I get one relation in $a,h$.
I wrote the tangent equation as $y=m(x-h)+a/m$. Putting $m=3/4$ and comparing with the given equation, I get another relation in $a,h$.
Solving simultaneously, I get $h=1/3,a=3/2$, thus focus is $(11/6,0)$
Therefore, the perpendicular distance from tangent is $5/2$.
Method $2$:
On doubtnut, they have a video solution.
They seem to be saying the slope of focal chord is $\tan2\theta$ if the slope of tangent is $\tan\theta$. Is this true?
Method $3$
Can we use the fact that the foot of perpendicular from focus on any tangent lies on the tangent at vertex?
If $T$ is where the tangent cuts the $x$-axis then will $FT$ equal $FP$? Yes.
Consider the points on the $x$-axis along with their $x$-coordinates $$T(t), D(h-a),V(h),F(h+a),K(r)$$ $D$ is the point where the directrix cuts the $x$-axis, $V$ and $F$ are the vertex and the focus respectively and $P(r,s)$ is the point on the parabola, therefore $FP=DK$. With this notation, the equation of the parabola becomes $y^2=4a(x-h)$ and we have $$s^2=4a(r-h)$$
The slope of the tangent line can be written in two ways, rise over run and also using the derivative
$$\frac {s}{r-t}=\frac {2a}{s}$$
Eliminate $s$ and $a$ from these equations to get
$$h=\frac{r+t}{2}$$
This shows that $V$ is the midpoint of $TK$ implying $FT=DK=FP$