Question 9
The equation of normal to the curve 3x2 โ y2 = 8 which is parallel to the line x + 3y = 8 is
(A) 3x โ y = 8 (B) 3x + y + 8 = 0
(C) x + 3y ยฑ 8 = 0 (D) x + 3y = 0
Since, the normal to the curve is parallel to the line ๐ฅ+3๐ฆ=8
โด Slope of normal = Slope of line
So, finding slope of normal and slope of line
Finding slope of normal
"3" ๐ฅ"2 โ" ๐ฆ"2 = 8"
Differentiating w.r.t. x
6๐ฅ โ 2๐ฆ ๐๐ฆ/๐๐ฅ = 0
2๐ฆ ๐๐ฆ/๐๐ฅ = 6๐ฅ
๐๐ฆ/๐๐ฅ =(6๐ฅ )/2๐ฆ
๐๐ฆ/๐๐ฅ =3๐ฅ/๐ฆ
Slope of normal =(โ1)/(๐๐ฆ/๐๐ฅ)
=(โ1)/(3๐ฅ/๐ฆ)
=(โ๐)/๐๐
Finding slope of line
๐ฅ+3๐ฆ=8
Differentiating w.r.t. x
1+3 ๐๐ฆ/๐๐ฅ = 0
๐๐ฆ/๐๐ฅ = (โ1)/3
Slope of line =๐๐ฆ/๐๐ฅ
=(โ๐)/๐
โด Equating (1) & (2)
(โ๐)/๐๐ = (โ๐)/๐
โ3๐ฆ=โ3๐ฅ
๐=๐
Now, to find equation of normal, we need a point
So, Putting ๐=๐ in the curve
3๐ฆ^2โ๐ฅ^2=8
3๐ฅ^2โ๐ฅ^2=8
2๐ฅ^2 = 8
๐ฅ^2 = 8/2
๐ฅ^2 = 4
๐=ยฑ๐
For y-coordinates, putting value of ๐ฅ in y=๐ฅ
Finding equation of normal
Equation of line at (๐ฅ_1, ๐ฆ_1) & having slope m is
(๐ฆโ๐ฆ_1 ) = m (๐ฅโ๐ฅ_1 )
For ๐ = 2
๐ฆ=๐ฅ
๐ฆ=2
So, the point is (2, 2)
For ๐ = โ2
๐ฆ=๐ฅ
๐ฆ=โ2
So, the point is (โ2, โ2)
Equation of normal at (2, 2) & Slope (โ๐)/๐
(๐ฆโ2) = (โ1)/3 (๐ฅโ2)
3 (๐ฆโ2) = โ1 (๐ฅโ2)
3๐ฆโ6=โ๐ฅ+2
3๐ฆ+๐ฅ=6+2
3๐ฆ+๐ฅ=8
๐๐+๐โ๐=๐
Equation of normal at (โ2, โ2) & Slope (โ๐)/๐
(๐ฆ+2) = (โ1)/3 (๐ฅ+2)
3 (๐ฆ+2) = โ1 (๐ฅ+2)
3๐ฆ+6=โ๐ฅโ2
3๐ฆ+๐ฅ=โ6โ2
3๐ฆ+๐ฅ=โ8
๐๐+๐+๐=๐
Hence, the required equation of normal is 3y + ๐ ยฑ 8 = 0
So, the correct answer is (C)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.