Characteristics of four quadrants

Consider the following figure-

The graph cuts x-axis at (4,0) and (-4,0). It cuts the y-axis at (0,4) and (0,-4).
The area enclosed is a square with vertices at the above mentioned points.
The number of points lying in Q1 is 6. The number of points on positive x-axis is 4.
Therefore, by symmetry, the number of points lying in all the four quadrants = 6*4 = 24.
The number of points lying on all four axes is 4*4 = 16. Plus the point (0,0).
Thus the total number of points enclosed by the graph = 41.

If you put -x in place of x and -y in place of y, the equation does not change => The curve is symmetrical in all the quadrants.
=> We can find the number of point in one quadrant and then multiply it by 4.
For now, let's omit the points on the axes.
$$\frac{x^2}{16}+\frac{y^2}{25}=1$$ can be written as $$25x^2+16y^2=400$$
For x = 1, y can take values from 1 to 4 => 4
For x = 2, y can take values from 1 to 4 => 4
For x = 3, y can take values from 1 to 3 => 3
For x = 4, y cannot take any values greater than 0
=> there are 11 points in each quadrant => 44 points in 4 quadrants
On the axes,

on x-axis i.e y=0.

==>$$25x^2+16y^2 \leq 400$$

==> $$x^2 \leq 16$$ ==> $$-4 \leq x \leq 4$$ ==> $$x \varepsilon {-4, -3, -2, -1, 0, 1, 2, 3, 4}$$ total 9 points.

on y-axis i.e x=0

==> $$25x^2+16y^2 \leq 400$$

==> $$y^2 leq 25$$ ==> $$-5 \leq y \leq 5$$ ==> $$y \varepsilon {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}$$ total 11 points.

Since orgin is included twice, total number of points = 9+11-1 = 19

=> 44 + 19 = 63 points in total.

Take a look at the given figure.

Since points A and B lie on the parabola $$y = x^2$$,

Let the coordinates of A be ($$-x_1,x_1^2$$) and B be ($$x_2,x_2^2$$).

AC = AB

Since the equation of line is y = x+b

Slope of the line = 1

This line makes an angle of 45 with x axis

In triangle ADC and BEC

Angle ADC = Angle BEC = 90
Angle C = Angle C which is common

Triangle ADC is similar to triangle BEC

AC/BC = AD/BE

AC/(AC+AB) = AD/BE

AD/BE = 1/2

$$x_1^2/x_2^2 = 1/2$$
$$x_2^2/x_1^2 = 2$$

Hence the answer is 2.

This question can also be solved using the following method,

As CA = AB

A is the mid- point of AB

Now AD is parallel to BE as AD and BE are perpendicular to the same line.

By mid point theorem

AD is half of BE

$$x_2^2/x_1^2 = 2$$