Let L be a normal to the parabola y² = 4x. If L passes through the point (9,6),then write the equation of L.
IiT-JEE 2011
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Answers
Given
The parabola y² = 4x
To Find
The equation of normal passing through (9,6).
Calculations
We know, normal to any parabola is given by y = mx - 2am - am³.
Here,for the given parabola y² = 4x.
4a = 4
=> a = 1.
Putting the value of a in the equation of normal.
y = mx - 2(1)m -(1)m³
=> y = mx - 2m - m³ .....(i)
Now,this normal passes through (9,6). Hence, putting x = 9 and y = 6,we get :-
=> 6 = 9m - 2m - m³
=> m³ - 7m - 6 = 0
=> m³ + m² - m² - m - 6m - 6 = 0
=> m²(m + 1) - m(m + 1) - 6(m + 1) = 0
=> (m + 1) ( m² - m - 6) = 0
=> (m + 1) (m+3) ( m - 2) = 0.
We get, m = -3, m = -2 and m = 2. Put this value in eqn(i) to get the equation of L.
putting m = -3
=> y = -3x - 2(-3) - (-3)³
=> y = -3x + 33 ......first equation of L
putting x = -1
=> y = - x -2(-1) -(-1)³
=> y = - x + 3 ....... second equation of L
putting x = 2
=> y = 2x - 2(2) - (2)³
=> y = 2x - 12....third equation of L.
Answer:
Given
The parabola y² = 4x
To Find
The equation of normal passing through (9,6).
Calculations
We know, normal to any parabola is given by y = mx - 2am - am³.
Here,for the given parabola y² = 4x.
4a = 4
=> a = 1.
Putting the value of a in the equation of normal.
y = mx - 2(1)m -(1)m³
=> y = mx - 2m - m³ .....(i)
Now,this normal passes through (9,6). Hence, putting x = 9 and y = 6,we get :-
=> 6 = 9m - 2m - m³
=> m³ - 7m - 6 = 0
=> m³ + m² - m² - m - 6m - 6 = 0
=> m²(m + 1) - m(m + 1) - 6(m + 1) = 0
=> (m + 1) ( m² - m - 6) = 0
=> (m + 1) (m+3) ( m - 2) = 0.
We get, m = -3, m = -2 and m = 2. Put this value in eqn(i) to get the equation of L.
putting m = -3
=> y = -3x - 2(-3) - (-3)³
=> y = -3x + 33 ......first equation of L
putting x = -1
=> y = - x -2(-1) -(-1)³
=> y = - x + 3 ....... second equation of L
putting x = 2
=> y = 2x - 2(2) - (2)³
=> y = 2x - 12....third equation of L.