Straight Lines

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0

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New answer posted•a year ago
S
a year ago

h = c o s θ + 3 2

           

=> k = s i n θ + 2 2

=> c o s θ = 2 h − 3 & s i n θ = 2 k − 2

( h − 3 / 2 ) 2 + ( k − 1 ) 2 = 1 4

∴ circle of radius r = 1 2

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Equation of normal is 4x – 3y + 1 = 0

Equation of tangent is 3x + 4y – 43 = 0

Area of triangle = 1 2 ( 4 3 3 + 1 4 ) × 7  

= 1 2 × ( 1 7 2 + 3 1 2 ) × 7 = 1 2 2 5 2 4          

  ∴ 2 4 A = 1 2 2 5        

 

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New answer posted•a year ago
S
a year ago

Let P (h, k)

( h − 5 ) 2 + k 2 = 3 ( h + 5 ) 2 + k 2           

h 2 − 1 0 h + 2 5 + k 2 = 9 h 2 + 9 0 h + 2 2 5 + 9 k 2           

  8 h 2 + 8 k 2 + 1 0 0 h + 2 0 0 = 0

x 2 + y 2 + 2 5 2 x + 2 5 = 0

r 2 = 6 2 5 1 6 − 2 5 = 6 2 5 − 4 0 0 1 6 = 2 2 5 1 6

4 r 2 = 2 2 5 4 = 5 6 . 2 5         

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New answer posted•a year ago
S
a year ago

y = ax2 + bx + c

a + b + c = 2

∴ d y d x = 2 a x + b ⇒ d y d x | ( 0 , 0 ) = b = 1    

It passes through (0, 0)

0 = 0 + 0 + c -> c = 0

a = 1

∴ a = 1 , b = 1 , c = 0           

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New answer posted•a year ago
S
a year ago

 x=y4, xy=k

dydx=14y3, dydx=−kx2

P (x1, y1)

where x1=y14&x1y1=k

⇒y1=k1/5, x1y1=k

m1m2=−1

⇒−14.k6/5=−1⇒k6/5=14⇒k6=145=12024

(4k)6=212.1210=22=4

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New answer posted•a year ago
S
a year ago

  a + 2 + a 3 = 1 0 3

2a + 2 = 0

2a = 8 -> a = 4       .(i)

and            c + b + b 3 = 7 3

2b + c = 7 .(ii)

Since         a, b, c are in A.P.

2b = a + c

From (i)   ∴ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2

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New answer posted•a year ago
S
a year ago

K = 4 3 3 x + y  

K =    3 x − y 4 3

∴ 4 3 3 x + y = 3 x − y 4 3   

x 2 1 6 − y 2 4 8 = 1

48 = 16 (e2 – 1)

e = 4 = 2      

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