Question 9 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 4, 2021 by Teachoo

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The equation of normal to the curve 3x
^{
2
}
β y
^{
2
}
= 8 which is parallel to the line x + 3y = 8 is

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Question 9
The equation of normal to the curve 3x2 β y2 = 8 which is parallel to the line x + 3y = 8 is
(A) 3x β y = 8 (B) 3x + y + 8 = 0
(C) x + 3y Β± 8 = 0 (D) x + 3y = 0
Since, the normal to the curve is parallel to the line π₯+3π¦=8
β΄ Slope of normal = Slope of line
So, finding slope of normal and slope of line
Finding slope of normal
"3" π₯"2 β" π¦"2 = 8"
Differentiating w.r.t. x
6π₯ β 2π¦ ππ¦/ππ₯ = 0
2π¦ ππ¦/ππ₯ = 6π₯
ππ¦/ππ₯ =(6π₯ )/2π¦
ππ¦/ππ₯ =3π₯/π¦
Slope of normal =(β1)/(ππ¦/ππ₯)
=(β1)/(3π₯/π¦)
=(βπ)/ππ
Finding slope of line
π₯+3π¦=8
Differentiating w.r.t. x
1+3 ππ¦/ππ₯ = 0
ππ¦/ππ₯ = (β1)/3
Slope of line =ππ¦/ππ₯
=(βπ)/π
β΄ Equating (1) & (2)
(βπ)/ππ = (βπ)/π
β3π¦=β3π₯
π=π
Now, to find equation of normal, we need a point
So, Putting π=π in the curve
3π¦^2βπ₯^2=8
3π₯^2βπ₯^2=8
2π₯^2 = 8
π₯^2 = 8/2
π₯^2 = 4
π=Β±π
For y-coordinates, putting value of π₯ in y=π₯
Finding equation of normal
Equation of line at (π₯_1, π¦_1) & having slope m is
(π¦βπ¦_1 ) = m (π₯βπ₯_1 )
For π = 2
π¦=π₯
π¦=2
So, the point is (2, 2)
For π = β2
π¦=π₯
π¦=β2
So, the point is (β2, β2)
Equation of normal at (2, 2) & Slope (βπ)/π
(π¦β2) = (β1)/3 (π₯β2)
3 (π¦β2) = β1 (π₯β2)
3π¦β6=βπ₯+2
3π¦+π₯=6+2
3π¦+π₯=8
ππ+πβπ=π
Equation of normal at (β2, β2) & Slope (βπ)/π
(π¦+2) = (β1)/3 (π₯+2)
3 (π¦+2) = β1 (π₯+2)
3π¦+6=βπ₯β2
3π¦+π₯=β6β2
3π¦+π₯=β8
ππ+π+π=π
Hence, the required equation of normal is 3y + π Β± 8 = 0
So, the correct answer is (C)

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