MHB Normals from a point to a parabola

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The discussion revolves around the properties of normals to the parabola defined by the equation y^2=4x. It is established that for the point (h,h+1) with h<2, there can only be one normal to the parabola, making Statement-1 true. Statement-2, which asserts that (h,h+1) lies outside the parabola for all h≠1, is also confirmed as true. However, participants conclude that Statement-2 does not adequately explain Statement-1, leading to the choice of option B. The analysis emphasizes the relationship between the cubic equation derived from the normal and the conditions for its roots.
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Problem:

STATEMENT-1: Through (h,h+1), there cannot be more than one normal to the parabola $y^2=4x$, if $h<2$.

STATEMENT-2: The point (h,h+1) lies outside the parabola for all $h\neq 1$.

A)Statement-1 is True, Statement-2 is True; Statement 2 is a correct explanation for Statement-1.

B)Statement-1 is True, Statement-2 is True; Statement 2 is NOT a correct explanation for Statement-1.

C)Statement-1 is True, Statement-2 is False.

D)Statement-1 is False, Statement-2 is True.

Attempt:

I figured out that the locus of the given point is $y=x+1$. I found that this equation is a tangent to given parabola, hence Statement-2 is certainly true.

I am unsure about how to proceed for Statement-1. Here's what I think:

The parametric coordinates of the given parabola is $(t^2,2t)$. The equation of normal in terms of $t$ is $y=-tx+2t+t^3$. As this normal passes through $(h,h+1)$, hence,

$$t^3+t(2-h)-(h+1)=0$$

But I am not sure how to proceed from here.

Any help is appreciated. Thanks!

EDIT: The title should be "Normals from a point to parabola". :p
 
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Re: Normals from a point to parabola

Pranav said:
The parametric coordinates of the given parabola are $(t^2,2t)$. The equation of normal in terms of $t$ is $y=-tx+2t+t^3$. As this normal passes through $(h,h+1)$, hence,

$$t^3+t(2-h)-(h+1)=0$$

But I am not sure how to proceed from here.
You want to know whether that cubic equation for $t$ has more than one solution. The cubic polynomial $f(t) = t^3+t(2-h)-(h+1)$ has either one real root or three real roots. If it has three roots then its graph must cross the axis three times. It would have to have a local maximum between the first two roots, and a local minimum between the second and third roots. So you could use calculus to find where (if anywhere) the turning points of $f(t)$ occur.
 
Re: Normals from a point to parabola

Hi Opalg!

Opalg said:
You want to know whether that cubic equation for $t$ has more than one solution. The cubic polynomial $f(t) = t^3+t(2-h)-(h+1)$ has either one real root or three real roots. If it has three roots then its graph must cross the axis three times. It would have to have a local maximum between the first two roots, and a local minimum between the second and third roots. So you could use calculus to find where (if anywhere) the turning points of $f(t)$ occur.

The question says that there should be only one normal i.e one real root. For the cubic to have only one real root, $f'(t)>0$ (f(t) must be strictly increasing) $\Rightarrow t^2>h/3$. I don't see what to do with the inequality I have got. :(
 
Re: Normals from a point to parabola

Pranav said:
For the cubic to have only one real root, $f'(t)>0$ (f(t) must be strictly increasing) $\Rightarrow t^2>h/3$. I don't see what to do with the inequality I have got. :(
That's not what I get from differentiating $f(t).$ In fact, $f'(t) = 3t^2 - (h-2).$
 
Re: Normals from a point to parabola

Opalg said:
That's not what I get from differentiating $f(t).$ In fact, $f'(t) = 3t^2 - (h-2).$

Sorry. :o

But the question still remains, how am I supposed to solve $t^2>(h-2)/3$? I am thinking that if h<2, the inequality always holds true but for any other h, we can always find a t which does not satisfy the inequality and hence, h<2 is the answer. Is this correct?
 
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Re: Normals from a point to parabola

Pranav said:
But the question still remains, how am I supposed to solve $t^2>(h-2)/3$? I am thinking that if h<2, the inequality always holds true but for any other h, we can always find a t which does not satisfy the inequality and hence, h<2 is the answer. Is this correct?
STATEMENT-1 is only concerned with what happens when $h<2$. You don't need to worry about what happens for any other $h.$
 
Re: Normals from a point to parabola

Opalg said:
STATEMENT-1 is only concerned with what happens when $h<2$. You don't need to worry about what happens for any other $h.$

Thanks Opalg! :)

I think I will go with B because I don't see how STATEMENT-1 can be deduced from STATEMENT-2.
 
Re: Normals from a point to parabola

Pranav said:
I think I will go with B because I don't see how STATEMENT-1 can be deduced from STATEMENT-2.
Nor do I. (Thinking)
 
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