If a focal chord with positive slope $m$ of the parabola $y^2 =16x$ touches the circle $x^2 +y^2-12x+34=0$ then the value of m is...?
I again use the result and I get equation of tangent of circle as (using result from page-91):
$$ Yy + X(x-6) + 68= 0$$
Or,
$$ Y + X \frac{x-6}{y} + \frac{68}{y}=0$$
Now, we can find equation of focal chord satisfying condition as $y=mx-4m$ or:
$$ y-mx+4m=0$$
Comparing,
$$ -m = \frac{x-6}{y}$$
And $$4m = \frac{68}{y}$$ But this leads to wrong answers!
I get $$x=23$$
Which is a point not even on the circle!
What is the mistake in my logic of applying this?
Equation of circle is $x^2 +y^2-12x+34=0$ or $(x-6)^2 + y^2 = 2$
The focal chord of the given parabola is $y = m(x-4)$ as you wrote.
The link in your question is not working for me but the easiest is to drop a perpendicular from center of the circle $(6, 0)$ to focal chord $y = m(x-4)$ and equate it to radius of the circle, $\sqrt2$ as the focal chord is tangent to the circle.
That gives $\frac{|0 - 6m + 4m|}{\sqrt{1+m^2}} = \sqrt2 \implies m = \pm 1$. As question says positive slope, we have $m = 1$.
Even by your method where it seems you are first finding equation of tangent to the circle and equating it to the focal chord,
Given the equation of the circle, taking derivative with respect to $x$, $2x + 2y y' - 12 = 0 \implies y' = \frac{6-x}{y}$
So tangent to circle at a given point $(x_0, y_0)$ is
$y - y_0 = \frac{6-x_0}{y_0} (x-x_0)$
But we also have $y = m (x - 4)$.
Solving them gives $x_0 = 5, y_0 = \pm1, m = \pm1$ and we consider $m = 1$.