Conic Sections - Online Test

Solving the above two equations we get,

x=acosα−bsinα---(i) 

y=asinα−bcosα ---(ii)

Squaring both sides of (i) and (ii) and adding both the equation we get  x2+y2=a2+b2.

which represents the locus of circle as

(i) it is quadratic equation

(ii) coeff of x2 = coeff of y2

(iiI) there is no trerm involving xy.

x2+y2+2gx+2fy+c=0 has center (-g,-f)

As normal passes through center so center will satisfy the equation of normal

put y=-f and x=-g in equation of the line y=mx+c

-f = m(-g) +c

which gives mg =f+c.

x2+y2=25

After completing the square,we get

(x−0)2+(y−0)2=52

so center is (0,0) and radius is 5 units.

As the normal should pass through the centre of the circle, center should satisfy the equation of normal

so 3x – 4y = 0 is the equation of normal as (0,0) satisfy this equation.

x = a cos θ + b sin θ , and y=asinθ−bcosθ

putting the value of x and y in x2+y2 

(acosθ+bsinθ)2+(asinθ−bcosθ)2  we get   a2+b2

=> x2+y2 =a2+b2

which is quadratic in nature,

coefficient of x2 = coefficient of y2

and no term involving xy

hence its locus represents a circle

The given lines are paralle. Hence the tangents are parallel tangents. Distance between these lines is equal to the diameter

D = |8−(−7)|36+64√ = 3/2

Therefore the radius is 3/4

Because line joining the points (a, 0) and (0, b)  will be diameter and it's mid-point will be centre of the circle.

Or let the equation of circle be (x−h)2+(y−k)2=r2

as circle passes through  (0,0) , (a,0) and (0,b), they will satisfy the above equation

putting the points in above equation we get

h2+k2=r2

(a−h)2+k2=r2  ---(ii)

h2+(b−k)2=r2 ---(iii)

Solving (i) and (ii),we get  h=a2

whereas solving (i) and (iii) we get k=b2

hence its center is (a2,b2)

x216+y29=1

comparing with the standard equation we get a=4 and b=3 we get c =16−9−−−−−√ =7–√

so focus is (7–√,0)  and the given centre is (0,3).

using distance formulae. we get radius =4 units