Test Level 3: Coordinate Geometry - CAT MCQ

For the given ellipse a² = 16 and b² = 9
⇒ a²(1 - e²) = 9
⇒ 16 - a²e² = 9
⇒ ae = √7
So, foci are at (- √7, 0) and (√7, 0).
Hence, required length of radius 

As the slope of AB is equal to the slope of DE, i.e. 1, so the equation of line AB is x = y ⇒ x - y = 0.

The vertex A (x, y) is equidistant from B and C because ABC is an isosceles triangle.
Therefore, (x - 1)² + (y - 3)² = (x + 2)² + (y - 7)²
⇒ 6x - 8y + 43 = 0
Now, from the options, we have
x = 5/6 and y = 6, which satisfy the above equation.

The diagrammatic representation of the same is as shown in the diagrams below.


In either case,

or a² = 4 or a = ± 2
As -2 is the only option available, a = -2
Hence, answer option 1 is correct.

Option (1):

4 and 5 are intercepts.
Option (2):

6 and 7 are intercepts.
Option (3):

4 and 3 are the intercepts.

Option (4):

19 and 10 are not consecutive.
So, option 4 is correct.

The perpendicular distance OP is given by the formula: 

Also, (x, y) = (0, 0)
⇒ k = ± 60

The four possible lines are:
x + y = 4; x - y = 4; - x - y = 4 and -x + y = 4
The four lines can be represented on the coordinate axes as shown in the figure.
Thus, a square is formed with the vertices as shown.
The side of the square is  

The area of the square is (4√2)² = 32 sq. units.

Slope of the line x + 3y = -7 is -1/3.
On solving the equation for the three lines, we get the vertices of the triangles as (-1, 2), (3, 5) and (1, 8).
So, the centroid of the triangle is ([-1 + 3 + 1] / 3, [2 + 5 + 8] / 3) or (1, 5) 
Using the slope-point form of the equation of a line, we get the required equation: y - 5 = (-1/3)(x - 1), i.e. x + 3y = 16.

The equations of parallel sides of the square are x + y − 1 = 0 and x + y + 2 = 0.
Therefore, Length of the side of the square = Distance between parallel side =

Hence, area of the square = (side)² = 9/2 sq.units

Given the points of triangle, A(p, 2 − 2p),B(1 − p, 2p) and C(−4 − p ,6 − 2p)
Area = 70sq.unit

{p[2p−6+2p]+(1−p)[6−2p+2p]+(−4−p)[2−2p−2p]}=140
p(4p−6)+(1−p)(4)−[(4+p)(2−4p)] =140
4p² −6p+4−4p−[8−16p+2p−4p²] =140
4p² −6p+4−4p−8+16p−2p+4p² =140
8p² +4p−144=0
2p_² +p−36=0
2p² +9p−8p−36=0
p(2p+9)−4(2p+9)=0
(2p+9)=0,(p−4)=0
p = −9/2,p = 4
∴ Only one integral value is possible.