How to solve this analytical geometry problem?-parable inscribed within a square

Question

This problem appeared on the network, and although it looks simple I am not sure of the result.

The polygon $ABCD$ is a square with side $4$ cm and the curve inscribed inside the square is a parabola, determine the length of the segment $AP$, if $QC$ is $3$ cm.

problem

Answer 1

The quadratic

• Passes through the coordinates $(0,0), (4, 3)$.
• Has a maximum at $X < 4$
• Has the form $ y = - A (x-X)^2+4$.

Can you take it from here? If not, explain what you've tried and where you're stuck.

$y = -\frac{9}{16}(x-\frac{8}{3})^2 + 4$.
Hence $AP = \frac{8}{3}$.

Answer 2

Hope you know differentiation to find maximum/minimum of standard parabola curve of form $y = ax^2+ bx + c $.

Taking B as origin since it passes through (0,0), we get c=0

$$ y =a x^2+bx \tag1 $$

Differentiate to get max tangent point

$$ 2 ax +b=0\to x_{t} =- b/(2a) \tag 2$$

plug into (1) and simplify

$$ y_t= -b^2/(4a) = 4 \text{ (given)} \tag3$$

Point $(4,3)$ is given lying on the parabola

$$ 3= 16 a +4 b \tag4 $$

Eliminate $a$ between (3) and (4) and simplify resulting in the quadratic equation

$$ b^2-4b+3=0,\; (b-1)(b-3)=0 ; b=(1,3);\;\tag5 $$

From(3) corresponding $a$ values are

$$ \left(\dfrac{-1}{16},\dfrac{-9}{16} \right) \tag 6$$

Plug $(a,b) $ values into (2) we get two values of tangent point

$$ x_t=(8,8/3) \tag7$$

The first value lies outside interval $0<x_t<4$ so is discarded, second value is taken. The parabola has equation

$$y=3 x\left(1-\dfrac{3x}{16} \right) \tag 8$$

whose graph is the one given in your question.