Question

(a) (4,4) is a point on a locus whose equation is y2 = kx. Show that (16,8) is another point on the locus.

(b) (2,3) is a point on a locus whose equation is 5x +by= 16. Show that (0,8) is another point of the locus​

Answer 1

Explanation:

plug the coordinate points into the equations and solve for the remaining variable.

b) (2,3) - 5(2) + b(3) = 16

10 + 3b = 16. 3b= 6. b = 2

(0,8) - 5(0) + 2(8) = 16. 0+ 16 = 16. ☆

a) (4,4) assume based on results you meant y squared. (y^2)

(4)^2 = 4k. 16 = 4k. k = 4

(8)^2 = 4(16). 64 = 64. ☆

Answer 2

Please check the explanation.

Explanation:

a)

Given

(4,4) is a point on a locus whose equation is y²=kx

Putting (4, 4) in y²=kx

(4)² = k(4)

16 = 4k

k = 16/4

k = 4

Checking the point (16, 8) on the locus

y²=kx

Putting (16, 8) and k =4 in y²=kx

y²=kx

(4)² = 4×4

16 = 16

L.H.S = R.H.S

Therefore, (16, 8) is another point on the locus.

b)

Given

(2, 3) is a point on a locus whose equation is 5x +by= 16

Putting (2, 3) in 5x +by= 16

5x+by=16

5(2)+b(3)=16

10+3b=16

3b=6

b=6/3

b=2

Putting (0, 8) and b=2 in 5x +by= 16

5x +by= 16

5(0)+2(8)=16

0+16=16

16=16

L.H.S = R.H.S

Therefore, (0, 8) is another point on the locus.