Parabola and tangents in analytic geometry

Question

A linear function which has a slope of $- \frac{4}{3} $ is intersects with

the parabola $y^2 = 8x$ at two points, $A$ and $B$. Prove that the tangent to the parabola at the point of $A$ and the tangent to the parabola at the point of $B$ are perpendicular.

EDIT1

Suggested (Narasimham)

INSTEAD OF

A linear function which has a slope of $- \frac{4}{3} $ is intersects with

SUBSTTUTE

*A straight line of slope $-\frac43 $ and passing through focus $ F \, (2,0)$ intersects *

My attempts: I've expressed the points in this way $$ A(\frac{a^2}{8},a) $$ $$B(\frac{b^2}{8},b)$$

Substituted them in the slope formula: $$m = \frac{y_1 - y_2}{x_1-x_2}$$ And equated it to the slope of the linear function from the beginning (-4/3). Then tried to use the fact that the multiplication of slopes of perpendicular linear functions equals to -1. Yet nothing lead me to answer. Will be grateful for help, thanks in advance :)

Answer 1

Your problem statement is incorrect. Your proposition is true for tangents at either end of a straight line (A,B) through the focus only.

Points $A,B$ are found by intersection of

$$ y^2 = 8 x,\quad, \frac{y-0}{x-2} = -\frac43$$

Find tangent equations at $A,B$ and check that they are perpendicular.

Answer 2

Using your formulation, the condition for the gradient to be $-\frac 43$ simplifies to become $$a+b=-6$$

However, the condition for the tangents to be perpendicular simplifies to become $$ab=-16$$

Therefore the statement is only going to be true for specific values of $a$ and $b$, and not generally true.