Question

Find points on the curve x^2/9+y^2/16=1 at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis

Answer 1

given curve , x²/9 + y²/16 = 1
differentiate with respect to x ,
2x/9 + 2y/16. dy/dx = 0
=> dy/dx = {-2x/9}/{2y/16} = -8x/9y


(i) parallel to x - axis .
then, slope of tangent of the curve is 0
e.g., dy/dx = -8x/9y = 0
=> x = 0
put x = 0 in the curve x²/9 + y²/16 = 1
0/9 + y²/16 = 1
=> 0 + y²/16 = 1
=> y² = 16
square root both sides,
y = ± 4
hence, points are (0, 4) and (0, -4)

(ii) parallel to y - axis
then, slope of tangent of the curve is 1/0
e.g., dy/dx = -8x/9y = 1/0
=> y = 0
put y = 0 in x²/9 + y²/16 = 1
so, x²/9 + 0/16 = 1
=> x² = 9
square root both sides,
x = ±3
hence, points are (3, 0) and (-3,0)

Answer 2

 gινєи ϲυяνє , ϰ²/⑨ + γ²/①⑥ = ① 
∂ιᏐᏐєяєиτιατє ωιτн яєѕρєϲτ το ϰ , 
②ϰ/⑨ + ②γ/①⑥. ∂γ/∂ϰ = ⓪ 
=> ∂γ/∂ϰ = {-②ϰ/⑨}/{②γ/①⑥} = -⑧ϰ/⑨γ 

(ι) ραяαℓℓєℓ το ϰ - αϰιѕ .
τнєи, ѕℓορє οᏐ ταиgєиτ οᏐ τнє ϲυяνє ιѕ ⓪ 
є.g., ∂γ/∂ϰ = -⑧ϰ/⑨γ = ⓪ 
=> ϰ = ⓪
ρυτ ϰ = ⓪ ιи τнє ϲυяνє ϰ²/⑨ + γ²/①⑥ = ① 
⓪/⑨ + γ²/①⑥ = ① 
=> ⓪ + γ²/①⑥ = ① 
=> γ² = ①⑥ 
ѕգυαяє яοοτ ϐοτн ѕι∂єѕ, 
γ = ± ④ 
нєиϲє, ροιиτѕ αяє (⓪, ④) αи∂ (⓪, -④) 

(ιι) ραяαℓℓєℓ το γ - αϰιѕ 
τнєи, ѕℓορє οᏐ ταиgєиτ οᏐ τнє ϲυяνє ιѕ ①/⓪ 
є.g., ∂γ/∂ϰ = -⑧ϰ/⑨γ = ①/⓪ 
=> γ = ⓪ 
ρυτ γ = ⓪ ιи ϰ²/⑨ + γ²/①⑥ = ① 
ѕο, ϰ²/⑨ + ⓪/①⑥ = ① 
=> ϰ² = ⑨
ѕգυαяє яοοτ ϐοτн ѕι∂єѕ, 
ϰ = ±③ 
нєиϲє, ροιиτѕ αяє (③, ⓪) αи∂ (-③,⓪).