Question

Find out the value of a if the point (-1,-2) lies on the locus.
mx² + y² - 2x + 4y + 17= 0

Answer 1

Answer:

Step-by-step explanation:

There is a rule in geometry.

If a point lies out of locus then if we plug in the point in locus we get a positive quantity.

for eg. f(x,y) = $ax^{2} + bx + c$

now if a point  lies outside the curve f(x,y) then

 $f(x_{1},y_{1})$ > 0

In this case

f(x,y) = $mx^{2} + y^{2} - 2x + 4y + 17 = 0$

If a point $(x_{1},y_{1})$ lies on the curve/locus then $f(x_{1},y_{1})$ = 0 and if the point lies inside the curve then  $f(x_{1},y_{1})$  < 0.

now f(-1,-2) = 0 = $m(-1)^{2} + (-2)^{2} - 2(-1) + 4(-2) + 17 = 0$

Thus,

m + 4 + 2 - 8 + 17 = 0

m = -15

Hope it helps!