Secant through the Focus
Given the following:
- ( p , f ( p ) ) and ( q , f ( q ) ) are points on the parabola f ( x ) = 4 a ( x − h ) 2 + k .
- ( p , f ( p ) ) , ( q , f ( q ) ) and ( h , k + a ) are collinear.
What is f ′ ( p ) × f ′ ( q ) ?
The answer is -1.
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2 solutions
Nice solution! It's nice that you only based on the slope. Is it obvious what h, k, and a are?
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I don't think it is obvious what if both ( p , f ( p ) ) and ( q , f ( q ) ) are on the same half of the parabola instead of in opposite halves.
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They will never be on the same half, because they are opposite ends of the secant passing through the focus.
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I have plotted the parabola with a = 4 1 , h = k = 1 . p can be 2 and q can be 4. It has nothing to do with secant. There are just two points on the parabola.
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@Chew-Seong Cheong – I don't know how you get you're plotting. But when h=1, k=1, a=0.25 and p=2, then q=0.75. Following the given.
I cannot add pictures in my comment. I cannot show you my plotting.
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@Emmanuel David – You are right. I forgot about fixing ( h , k + a ) on the graph.
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@Chew-Seong Cheong – It's alright Glad that we've come to a common point hahahahahaha pun intended. XD
@Chew-Seong Cheong – Here is the graph where h=1, k=1, a=0.25 and p=2 https://www.desmos.com/calculator/ovljeektni
As p approaches infinity, f(q) will only approach k.
(h, k+a) is the focus. (p, f(p) and (q, f(q) are the endpoints of the secant going through the focus. And the tangent lines at the endpoints of a secant going through the focus are perpendicular and intersect on the directrix (This is true for all parabolas). Perpendicular slopes are negative reciprocal of each other. So their product is -1.
Here's an interactive graph that I prepared:
https://www.desmos.com/calculator/asfosixi23
Given that f ( x ) = 4 a ( x − h ) 2 + k , ⟹ f ′ ( x ) = 2 a x − h , ⟹ f ′ ( p ) f ′ ( q ) = 4 a 2 ( p − h ) ( q = h ) .
Since ( p , f ( p ) ) , ( q , f ( q ) ) and ( h , k + a ) are collinear, we have:
p − h f ( p ) − ( k + a ) p − h 4 a ( p − h ) 2 + k − ( k + a ) p − h ( p − h ) 2 − 4 a 2 ( p − h ) 2 ( q − h ) − 4 a 2 ( q − h ) ( p − h ) 2 ( q − h ) − ( q − h ) 2 ( p − h ) ( p − h ) ( q − h ) ( p − q ) 4 a 2 ( p − h ) ( q − h ) f ′ ( p ) f ′ ( q ) = q − h f ( q ) − ( k + a ) = q − h 4 a ( q − h ) 2 + k − ( k + a ) = q − h ( q − h ) 2 − 4 a 2 = ( q − h ) 2 ( p − h ) − 4 a 2 ( p − h ) = 4 a 2 ( q − h ) − 4 a 2 ( p − h ) = 4 a 2 ( q − p ) = p − q q − p = − 1