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प्रश्न
Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 = 145 at the point (x1, y1), where x1 = 2 and y1 > 0.
उत्तर
Given curve 16x2+9y2=145 ... (i)
Putting x1=2 in (i), we get
`16(2)^2 + 9y^2 = 145`
`64 + 9y^2 = 145`
`9y^2 = 81 `
`y^2 = 9`
`y = +-3`
Given that y > 0 so y = 3
`y^2 = (145-16x^2)/9` From i
Differentiating above both sides with respect to x
2y (dy)/(dx)
` = 1/9 (0- 32x)`
`(dy)/dx = (-32x_1)/(18y)`
`((dy)/(dx))_((2 "," 3)) = (-64)/54 = (-32)/27`
`:. (dy/dx)_((2","3)) = -32/27`
The equation of the tangent at (2,3) is
`y - 3 = (dy/dx)_(2","3) (x - 2)`
`y - 3 = (-32)/27 (x - 2)`
`=> 27y - 81 = -32x + 54`
`=> 32x + 27y - 135 = 0`
Now, the normal to the curve at (x1, y1) will be perpendicular to the tangent to the curve at (x1, y1)
Let normal to the curve have the slope m1
Then `m_1 xx((-32)/27) = -1`
`:.m_1 = (27/32)`
The equation of the normal at (2,3) is
`y - 3 = (27/32) (x - 2)`
`=> 32y−96=27x−54`
`=> 27x - 32y - 42 = 0`
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