Question 7 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

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

(A) 3x – y = 8 (B) 3x + y + 8 = 0

(C) x + 3y ± 8 = 0 (D) x + 3y = 0

Transcript

Question 7
The equation of normal to the curve 3x2 – y2 = 8 which is parallel to the line x + 3y = 8 is
(A) 3x – y = 8 (B) 3x + y + 8 = 0
(C) x + 3y ± 8 = 0 (D) x + 3y = 0
Since, the normal to the curve is parallel to the line 𝑥+3𝑦=8
∴ Slope of normal = Slope of line
So, finding slope of normal and slope of line
Finding slope of normal
"3" 𝑥"2 –" 𝑦"2 = 8"
Differentiating w.r.t. x
6𝑥 − 2𝑦 𝑑𝑦/𝑑𝑥 = 0
2𝑦 𝑑𝑦/𝑑𝑥 = 6𝑥
𝑑𝑦/𝑑𝑥 =(6𝑥 )/2𝑦
𝑑𝑦/𝑑𝑥 =3𝑥/𝑦
Slope of normal =(−1)/(𝑑𝑦/𝑑𝑥)
=(−1)/(3𝑥/𝑦)
=(−𝒚)/𝟑𝒙
Finding slope of line
𝑥+3𝑦=8
Differentiating w.r.t. x
1+3 𝑑𝑦/𝑑𝑥 = 0
𝑑𝑦/𝑑𝑥 = (−1)/3
Slope of line =𝑑𝑦/𝑑𝑥
=(−𝟏)/𝟑
∴ Equating (1) & (2)
(−𝒚)/𝟑𝒙 = (−𝟏)/𝟑
−3𝑦=−3𝑥
𝒚=𝒙
Now, to find equation of normal, we need a point
So, Putting 𝒚=𝒙 in the curve
3𝑦^2−𝑥^2=8
3𝑥^2−𝑥^2=8
2𝑥^2 = 8
𝑥^2 = 8/2
𝑥^2 = 4
𝒙=±𝟐
For y-coordinates, putting value of 𝑥 in y=𝑥
Finding equation of normal
Equation of line at (𝑥_1, 𝑦_1) & having slope m is
(𝑦−𝑦_1 ) = m (𝑥−𝑥_1 )
For 𝒙 = 2
𝑦=𝑥
𝑦=2
So, the point is (2, 2)
For 𝒙 = −2
𝑦=𝑥
𝑦=−2
So, the point is (−2, −2)
Equation of normal at (2, 2) & Slope (−𝟏)/𝟑
(𝑦−2) = (−1)/3 (𝑥−2)
3 (𝑦−2) = −1 (𝑥−2)
3𝑦−6=−𝑥+2
3𝑦+𝑥=6+2
3𝑦+𝑥=8
𝟑𝒚+𝒙−𝟖=𝟎
Equation of normal at (−2, −2) & Slope (−𝟏)/𝟑
(𝑦+2) = (−1)/3 (𝑥+2)
3 (𝑦+2) = −1 (𝑥+2)
3𝑦+6=−𝑥−2
3𝑦+𝑥=−6−2
3𝑦+𝑥=−8
𝟑𝒚+𝒙+𝟖=𝟎
Hence, the required equation of normal is 3y + 𝒙 ± 8 = 0
So, the correct answer is (C)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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