A)
sin−1 x +
sin−1 y =
sin−1 (
x√1−y2+y√1−x2) simplify to
get ∠C = π/2 then cot
+ cot
=
and cot
. cot
=
+ =
− =
→ cot
(+) = 1 on simplification , we get p – r = q.
∴ px + qy + r = 0 → px + qy + p – q = 0 → p(x + 1) + q(y – 1) = 0.
It passes through (–1, 1).
B. Using formula find images to get B ≡ (1, 8) and C ≡ (7, 2). A is (3, 4).
Circumcentre is (8, 9).
C. Any normal to the hyperbola xy = 1 is
xt3 – yt –
t4 + 1 = 0.
If it passes through (h, k) then
t4 –
ht3 + kt – 1 = 0.
Co-normal points are ti, 1/ti). Then
t1+t2 +
t3+t4 = h or
x1+x2+x3+x4 = h
and 1/
t1 + 1/
t2 + 1/
t3 + 1/
t4 = k or
y1+y2+y3+y4 = k
Let the variable line bc ax + by + c = 0.
Given that a
+b + c = 0 or a
+b + c = 0
∴ It passes through
(,) Here (h, k) = (32, 36).
D) f (x) =
when x ≠ 0 , is constant x = 0
∴
f(x) should exists and finite
is finite using L. Hosp value
→ 4 + b = 0 or b = - 4
then
=
→ a = - 2
substitute these values and final limit which is 2.
∴ f(0) =
f(x) = 2
|z + 1 – zi| = Im z → |z – (–1 + zi)| = Im z is a parabola whose focus is (–1, 2)
and directrix. y = 0.
∴ vertex is (–1, 1).