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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 :)

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    Which lines are you required to show are perpendicular? Can you express their slopes in terms of $(x_1,y_1)$ and $(x_2,y_2)$? – 2017-01-11
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    @MatthewLeingang the lines required to show that perpendicular are: 1- the tangent to the parabola at the point of A, 2- the tangent to the parabola at the point of B – 2017-01-11
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    And yeah, i can, that's what I've done in some way – 2017-01-11
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    Good; add that into your post. You have two equations in the two unknowns $y_1$ and $y_2$ (or $a$ and $b$ as you have them). – 2017-01-11
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    What is the slope of the tangent to the parabola at a given point $(x_0,y_0)$? – 2017-01-11
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    @MatthewLeingang as i see it i have one equation with two unknowns. Can't really assume initially that their multiplication of slopes equals to -1. – 2017-01-11
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    @amd I think that is a good question, and I would further suggest writing a condition for the tangents at two points on the parabola to be perpendicular. – 2017-01-11
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    @amd its $\frac{4}{y_0}$ – 2017-01-11
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    Hint: In your expression for $m$, what happens if you substitute for $x_1, x_2$ in the denominator using the equation of the parabola? – 2017-01-11
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    @Ozk You're right; it's more like three equations in four unknowns. You have $y_1^2 = 8x_1$, $y_2^2 = 8x_2$, $\frac{y_1-y_2}{x_1-x_2} = - \frac{4}{3}$, and you want to show $y_1 y_2 = -16$. – 2017-01-11
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    @ Ozk Tangents at a focal chord ends are orthogonal, cut on the directrix,a property of parabola. – 2017-01-11
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    @Narasimham Seems to me that the problem at hand is essentially to prove this property in a specific case. – 2017-01-11
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    Seems like there’s something missing from the problem statement. If you take the line $4x+3y=0$, one point of intersection is the origin, where the tangent to the parabola is vertical. That means that the tangent at the other point of intersection must be horizontal, which is impossible. – 2017-01-11

2 Answers 2

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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.

ParabolaTgts

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    So are you saying that it might be a problem with the question itself? – 2017-01-11
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    Yes, see my answer – 2017-01-11
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    i know that they're interchangeable but how it is leading me to a solution? – 2017-01-12
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    I have edited your question in the way suggested. If not OK, feel free to roll it back. – 2017-01-12
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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.

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    That's right but i cant assume that $ab = -16$ initially, i have to conclude it by substitution. – 2017-01-12