Answer:
11. Equation of hyperbola having vertices at (0, ±6) and foci at (0, ±9).
is given by ![\frac{y^2}{a^2}- \frac{x^2}{b^2}=1](https://tex.z-dn.net/?f=%5Cfrac%7By%5E2%7D%7Ba%5E2%7D-%20%5Cfrac%7Bx%5E2%7D%7Bb%5E2%7D%3D1)
also b²= c²-a²
b²=81-36
b²=45
So, Equation becomes , ![\frac{y^2}{36}- \frac{x^2}{45}=1](https://tex.z-dn.net/?f=%5Cfrac%7By%5E2%7D%7B36%7D-%20%5Cfrac%7Bx%5E2%7D%7B45%7D%3D1)
Option (D) is correct.
12.Vertices (0, ± 4), Asymptotes = ±1/4.x
equation of asymptote is given by, ![x = \pm y \frac{b}{a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cpm%20y%20%5Cfrac%7Bb%7D%7Ba%7D)
![\frac{4}{b}=\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7Bb%7D%3D%5Cfrac%7B1%7D%7B4%7D)
So a=4 and b=16,
So , equation becomes ![\frac{y^2}{16}- \frac{x^2}{256}=1](https://tex.z-dn.net/?f=%5Cfrac%7By%5E2%7D%7B16%7D-%20%5Cfrac%7Bx%5E2%7D%7B256%7D%3D1)
Option (B) is correct.
13. x= t-3, and y = t²+ 5
Replace t by x+3, we get
y= (x+3)² +5
y=x²+ 6x +14
Option (A) is correct.
14. Polar coordinates is given by (r,∅)
Polar coordinates of point is (3, 2π/3)
So, r =3, ∅ =2π/3
x= r cos∅ and y = r sin∅
x=3 Cos (2π/3) and y= 3 Sin (2π/3)
x= -3/2 and y=3√3/2
Option (A) is correct.
14. Polar coordinate is given by (r,∅)
Here , r=1, and ∅ = -π/6
x= r Cos ∅ and y =r Sin∅
x= 1 Cos (-π/6) and y= 1 Sin (-π/6)
x =√3/2 and y =1/2
Option (B) which is B) (1, negative pi divided by 6 + 2nπ) or (-1, negative pi divided by 6 + 2nπ) is correct.
16. Two pairs of polar coordinates for the point (4, 4) with 0° ≤ θ < 360°.
is option (B) which is B) (4 square root 2 , 45°), (-4 square root 2 , 225°).
17. A circular graph with inner loop on the left of a limacon curve is given by
r = a + b Cos∅.
In this case a> b.
So , Option (D) r = 4 + cos θ as well as (A) r = 3 + 2 cos θ looks correct.
18. Equation of limacon curve is given by r = -2 + 3 cos θ, here , a<b
So it is symmetric about y-axis only.Option (B) is correct.