The figure above shows the graph of y = a − x^2 for some constant a.

Question

https://gmatclub.com/forum/download/file.php?id=47432 The figure above shows the graph of y = a − x^2 for some constant a. If the square ABCD intersects the graph at points A and B and the area of the square is 16, what is the value of a ? (A) 2 (B) 4 (C) 6 (D) 8 (E) 10 2019-04-17_1143.png

Prateek1793 · Answer

Since the pt. A and B lie on the graph, it means these points should satisfy the equation y = a-x^2.
Thus, the y coordinate of both A and B is 4. Why? Since the square area is 16, therefore the side length is 4.
Thus, Pt A is (-2,4) and Pt B is (2,4). 
Putting these values in the above equation y= a-x^2, we get a = 8.

Hence, D

Timebomb · Answer

The distance between points A and B is 4 units. Let the x coordinte of point A be X and that of Point B X+4.

Putting it in equation of parabola as both points satisfy the equation and remember that both points have y co-ordinate as 4. Moreover the point is that the y co-ordinate for both the points is same. SUbstituting

a- x^2 = a- (X+4)^2
Solving this you will get X = -2.

The point A therefore becomes ( -2,4) and B becomes (2,4) .
Putting any one of these in the equation will lead to

4= a-x^2
4=a- 2^2
a=8
Option C.

CrackverbalGMAT · Answer

This is an excellent question which tests you on multiple concepts – quadratic equations, shifting of graphs and co-ordinate geometry. So, let’s quickly review some of these concepts.

The standard graph for a quadratic equation of the form a\(x^2\) + bx + c = 0, is a parabola whose axis is a vertical line . When a>0, we get an upright parabola, whereas, when a<0, the parabola is inverted.
So,the standard graph of \(x^2\) = 0 is a parabola whose roots are the co-ordinates of the origin. Therefore, -\(x^2\) = 0 is an inverted parabola with the same roots as above.

If we need the graph of f(x) + a, where a is a constant, the graph of the function f(x) is to be shifted a units upwards. Similarly, the graph of f(x) – a can be obtained by shifting the graph of f(x), a units downwards.
The graph given to us is y = a – \(x^2\). We can rewrite this as –\(x^2\) + a. This means, the graph of y= -\(x^2\) has been shifted ‘a’ units upwards, which is corroborated by the diagram as well.

If any point lies on a line/curve, the co-ordinates of the point will satisfy the equation of the line/curve on substitution.

We know that the area of the square is 16 units. Therefore, AB = BC = 4 units.

Let the co-ordinates of point B be (p,q). Knowing that BC = 4 units, we can conclude that q = 4. So, the co-ordinates of B can now be written as (p,4).

Since B and C are on the same vertical line, their x co-ordinates will be the same. Since C is ON the x-axis, we can conclude that the co-ordinates of C are (p,0).

Let the co-ordinates of D be (r,0). Then, CD = √\((p-r)^2\).

But CD = AB = 4.

Therefore, √\((p-r)^2\) = 4 which means (p-r) = 4. This equation can be satisfied in many ways. For example:
p = 2 and r = -2 OR

p = 4 and r = 0.

But, what is to be noted is that y = -\(x^2\) is a parabola with y-axis as its axis. If we take the case of p = 4 and r = 0, we are taking one side of the square to be lying on the y-axis and the other outside the parabola. In this case, the vertices A and B will not be lying on the curve. Something like this:

Attachment:

21st June 2019 - Reply 1.JPG [ 17.91 KiB | Viewed 10835 times ]

So, we can say for sure that p = 2 and r = -2 is the only combination that lets A and B be on the curve. Now, we can say that the co-ordinates of B are (2,4).

Substituting this in the equation of the curve, we have,
4 = -\((2)^2\) + a
Therefore, a = 4 + 4 = 8.
The correct answer option is D.

Hope this helps!

Kinshook · Answer

Given: The figure above shows the graph of y = a − x^2 for some constant a. 
Asked: If the square ABCD intersects the graph at points A and B and the area of the square is 16, what is the value of a ?

Coordinate of point D: 
x = 2
y = 2x = 4
so that square has area = 16

Graph of y = a - x^2 at point D: 
y = a - 2^2 = 4; a = 8

IMO D

Advait01 · Answer

Since the pt. A and B lie on the graph, it means these points should satisfy the equation y = a-x^2.
Thus, the y coordinate of both A and B is 4. Why? Since the square area is 16, therefore the side length is 4.
Thus, Pt A is (-2,4) and Pt B is (2,4).
Putting these values in the above equation y= a-x^2, we get a = 8.

Hence, D

The figure above shows the graph of y = a − x^2 for some constant a.

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The figure above shows the graph of y = a − x^2 for some constant a.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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