If the distance between the two
6.
The point to which (0,0) is to be shifted to eliminate x and y terms of the equation 4x2+9y2
8r+360+4-0 is
1) (1, 3)
2) (4, 3)
3) (-1,2)
4) (1, -2)
Answers
Answer:
answer is -1, 2
Step-by-step explanation:
MATHS
A normal to the hyperbola, 4x
2
−9y
2
=36 meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram OABP(O being the origin) is formed, then the locus of P is
ANSWER
given hyperbola is: 4x
2
−9y
2
=36
let (x
0
,y
0
) be point of contact of normal on the hyperbola
Finding slope of normal at that point:
Differntiating hyperbola equation we get; 4×2×x−9×2×y
dx
dy
=0
⇒
dx
dy
=
9y
4x
= slope of tangent
∴ slope of normal=
4x
−9y
equation of normal at (x
0
,y
0
) is:
y−y
0
=
4x
0
−9y
0
(x−x
0
)
line intersects X axis at A when y=0
∴A=(
9
13x
0
,0)
similarly B=(0,
4
13y
0
)
given OABP forms a paralleogram→diagonals bisect each other(midpoint of diagonals are same)
midpoint of OB=(0,
8
13y
0
)=midpoint of AP
Let P=(x,y)
∴midpointofAP=
⎝
⎜
⎜
⎛
2
9
13x
0
+x
,
2
y
⎠
⎟
⎟
⎞
∴P(x,y)=(
9
−13x
0
,
4
13y
0
)→1
As (x 0 ,y 0 ) lie on hyperbola,it shoud the its equation:
4(x 0 )2
−9(y
0
)
2
=36
from equation 1: x
0
=
13
−9x
and y
0
=
13
4y
substituting in hyperbola equation ,we get:
9x
2
−4y
2
=169
∴ locus of point P is hyperbola whose equation is:9x
2
−4y
2
=169
hence correct option is C.
hope it helps you