Post mathematic related questions here

OP
rst

rst

Youngling
If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option

(A)16 ( a^2 + b^2) = c ^2
(B)16 ( a^2 - b^2) = c ^2
(C)16 ( a^2 +b^2) = - c^2
(D)16 ( a^2 - b^2) = - c^2
 
nomad47

nomad47

Cyborg Agent
rst said:
If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option

(A)16 ( a^2 + b^2) = c ^2
(B)16 ( a^2 - b^2) = c ^2
(C)16 ( a^2 +b^2) = - c^2
(D)16 ( a^2 - b^2) = - c^2
Click to expand...

option A.
As the line is a tangent there will be a single solution for the set of equation. Substitute either x or y from the linear equation in the quadratic equation and use condition both the roots are equal. You will arrive at A.
 
OP
rst

rst

Youngling
nomad47 said:
option A.
As the line is a tangent there will be a single solution for the set of equation. Substitute either x or y from the linear equation in the quadratic equation and use condition both the roots are equal. You will arrive at A.
Click to expand...

Answer is correct

Roots are equal i.e D=0

But it will form equation of degree 4 either in a or b

How will we get option (A)
 
Last edited:
OP
rst

rst

Youngling
rst said:
If ax + by = c is tangent to the circle x^2 + y^2 = 16 then which of the following is correct option

(A)16 ( a^2 + b^2) = c ^2
(B)16 ( a^2 - b^2) = c ^2
(C)16 ( a^2 +b^2) = - c^2
(D)16 ( a^2 - b^2) = - c^2
Click to expand...

x^2 + y^2 = 16

Here centre =(0,0)
r=4

Perpendicular distance from centre (0,0) to the tangent = r

c / sq rt ( a^2 + b^2) =4

squaring both sides, we get

16 ( a^2 + b^2) = c ^2
 
nomad47

nomad47

Cyborg Agent
rst said:
x^2 + y^2 = 16

Here centre =(0,0)
r=4

Perpendicular distance from centre (0,0) to the tangent = r

c / sq rt ( a^2 + b^2) =4

squaring both sides, we get

16 ( a^2 + b^2) = c ^2
Click to expand...

That's one way to do that. And am really bad with all these formulas. :p
 
OP
rst

rst

Youngling
The distance between the directrices of the ellipse 9 x^2 + 4 y^2 = 36 is which of the following option
(A) 2 √5
(B) √5
(C) 9/√5
(D) 18/ √5
 
OP
rst

rst

Youngling
rst said:
The distance between the directrices of the ellipse 9 x^2 + 4 y^2 = 36 is which of the following option
(A) 2 √5
(B) √5
(C) 9/√5
(D) 18/ √5
Click to expand...

9 x^2 + 4 y^2 = 36

(x^2) /4 + (y^2 )/9 = 1

Here a^2 =9, b^2 =4

a=3 , b= 2

As a^2 = b^2 + c^2
9= 4 + c^2
√5 =c
√5 =a e
√5/3 =e

The distance between the directrices =2a/e
= 18/ √5
 
OP
rst

rst

Youngling
powerhoney said:
Hey, how to find limit of ((2^x-1)/x)^2 as x tends to zero???
Click to expand...

lim (x->0) ((2^x-1)/x)^2

lim (x->0) (2^x-1)^2 / x^2

lim (x->0) (2^2x+1-2.2^x ) / x^2
Using L' Hospital rule

lim (x->0) (2 log2 2^2x -2.2^x log2) / 2x
lim (x->0) ( log2 2^2x -2^x log2) / x

Using L' Hospital rule

lim (x->0) [ 2 (log2)^2 2^2x -2^x (log2)^2]/ 1

=(log2)^2
 
Last edited:
OP
rst

rst

Youngling
powerhoney said:
Find the total number of squares and rectangles in an NxN chessboard
Click to expand...

In 8x8 chessboard we have only ONE 8x8 square.
There are FOUR 7x7 squares. There are NINE 6x6 squares, and so on

In 8x8 chessboard we have 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 square

Hence in NxN chessboard chessboard
the total number of squares = 1^2 + 2^2 + 3^2 + ... + N^2

Also NxN chessboard chessboard
the total number of rectangle = [C(n+1,2) ] ^2
 
You must log in or register to reply here.
Top Bottom