Conic Sections - Online Test

Answer : Option C
Explaination / Solution:

putting the value y=c into parabola,we get

x2=c-1

or x2-(c-1)=0

here discriminat=4c−4−−−−−√

line y=c is tangent when discriminant is equal to 0.

putting disriminant =0 we get c=1.

(0, c) will be a point on the parabola. 

Answer : Option C
Explaination / Solution:

Given centre = (1,2)

Therefore radius = (4−1)2+(6−2)2−−−−−−−−−−−−−−−√ = 5 units

Area of the circle is πr2 = 25π sq units

Answer : Option A
Explaination / Solution:

Let the point be (x,y)

Hence (x-2)2 + (y-0)2 = 23(x−9/2)1√2

(x- 2)2 + y2 = 23(x - 9/2)

On simplifying we get the equation of an ellipse

Answer : Option D
Explaination / Solution:

Answer : Option A
Explaination / Solution:

Since the value of k is not specified, we cannot get a specific equation.

Answer : Option C
Explaination / Solution:

x2−y2=9

above equation can be written as,

x2(3)2−y2(3)2=1

comparing it with the standard equation we get a=3 and b=3

as c=a2+b2−−−−−−√

we get c= 3 2–√

and as e = ca

we get e = 2–√

Answer : Option A
Explaination / Solution:

After solving the equations we will get x+y = 2sec θ which represents a linear equation.

Answer : Option D
Explaination / Solution:

y = m x + c ---(i)

y2=4ax ---(ii)

putting the value of y from (i) in (ii), we get

(mx+c)2=4ax

=> (mx)2+x(2mc−4a)+c2=0

here

discriminant =(2mc−4a)2−4m2c2−−−−−−−−−−−−−−−−−√

                     =16a2−16amc−−−−−−−−−−−√

when discriminant >0 line touches parabola at two points,

when discriminant < 0 line do not  touches parabola and

when discriminant = 0 line touches parabola at one point

and we know that tangent is a line that touches the curve at exactly one point

so putting discriminat = 0 and solving

we get c=am,m≠0

 

 

on putting the value of y from line in the parabola and solving for equal roots.

Answer : Option B
Explaination / Solution:

y=4at

Squaring both sides,we get

y2=16a2t2

Putting the value of at2 i.e. at2=y216a in  x = at2 we get,

16ax=y2

or y2=4(4a)x

which is nothing but equation of parabola.

Answer : Option B
Explaination / Solution:

x=et+e−t2,y=et−e−t2;t∈R

Squaring both sides of both the equation ,we get

$$x2 = (et+e−t)24 and   y2 = (et−e−t)4

Subtracting one equation from another we get

x2 - y2 = 1 which is nothing but equation of hyperbola