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Question : 48 of 160
Marks:
+1 ,
-0
Solution:
(*) Lines represented by
4 x 2 − 5 x y + 3 y 2 = 0 are
L 1 and
L 2 .
3 ( ‌ ) 2 − 5 ( ‌ ) + 4 ‌ = 0 3 m 2 − 5 m + 4 ‌ = 0 < m 1 m 1 + m 2 ‌ = ‌ , m 1 ⋅ m 2 ‌ = ‌ ‌ + ‌ ‌ = ‌ = ‌ Let the slope of line
L 1 is
m 1 and the slope of line
L 2 is
m 2 .
L 3 is perpendicular to
L 1 ⇒ So, slope of
L 3 = − ‌ L 4 is perpendicular to
L 2 ⇒ So, slope of
L 4 = − ‌ Let the equation of
L 3 : y = ‌ x + c 1 It passes through
( 4 , 3 ) : 3 = ‌ + c 1 ⇒
c 1 = 3 + ‌ L 3 : y = ‌ x + 3 + ‌ ⇒ y + ‌ − ( 3 + ‌ ) = 0 and Equation of
L 4 : y = − ‌ + c 2 Passes through
( 4 , 3 ) : 3 = − ‌ + c 2 ⇒ ‌ ‌ c 2 = 3 + ‌ L 4 : y = − ‌ + 3 + ‌ ⇒ y + ‌ − ( 3 + ‌ ) = 0 Combined equation of
L 3 and
L 4 [ y + ‌ − ( 3 + ‌ ) ] [ y + ‌ − ( 3 + ‌ ) ] = 0 ⇒ y 2 + ‌ − y ( 3 + ‌ ) + ‌ + ‌ − ‌ ( 3 + ‌ ) − y ( 3 + ‌ ) − ‌ ( 3 + ‌ ) + ( 3 + ‌ ) ( 3 + ‌ ) = 0 ⇒ y 2 + ‌ + x y ( ‌ + ‌ ) − x ( ‌ + 3 ( ‌ + ‌ ) ) − y ( 6 + 4 ( ‌ + ‌ ) ) + ( 9 + 12 ( ‌ + ‌ ) + ‌ ) = 0 ⇒ y 2 + ‌ x y + ‌ x 2 − x ( 6 + 3 × ‌ ) − y ( 6 + 4 × ‌ ) + ( 9 + 12 × ‌ + 16 × ‌ ) = 0 ⇒ y 2 + ‌ x y + ‌ x 2 − ‌ x − 11 y + 36 = 0 ⇒ 4 y 2 + 5 x y + 3 x 2 − 39 x − 44 y + 144 = 0 On comparing with
a x 2 + 2 h x y + b y 2 + 2 g x + 2 f y + c = 0 a = 4 , h = ‌ , b = 3 , g = ‌ , f = − 22 and
c = 144 a f + b g + c h = − 88 − ‌ + 360 = 2135 © examsiri.com
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