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प्रश्न
Find the equations of tangents and normals to the following curves at the indicated points on them : `x = sqrt(t), y = t - (1)/sqrt(t)` at = 4.
उत्तर
When t = 4, x = `sqrt(4) and y = 4 - (1)/sqrt(4)`
∴ `x = 2 and y = 4 - (1)/(2) = (7)/(2)`
Hence, the point at which we want to find the equations of tangent and normal is `(2, 7/2)`.
Now, `x = sqrt(t), y = t - (1)/sqrt(t)`
Differentiating x and y w.r.t. t, we get
`dx/dt = d/dt(sqrt(t)) = (1)/(2sqrt(t))`
and `dy/dx = d/dt(t - 1/sqrt(t))`
= `1 / (-1/2)t^(-3/2)`
= `1 + (1)/(2t^(3/2)`
= `(2t^(3/2) + 1)/(2t^(3/2)`
∴ `dy/dx = ((dy/dt))/((dx/dt)`
= `(((2t^(3/2 + 1))/(2t^(3/2))))/((1/(2sqrt(t)))`
= `(2t^(3/2) + 1)/(2t^(3/2)) xx 2sqrt(t)`
= `(2t^(3/2) + 1)/t`
∴ `(dy/dx)_("at" t = 4) = (2(4)^(3/2) + 1)/(4)`
= `(2 xx 8 + 1)/(4)`
= `(17)/(4)`
= slope of the tangent at t = 4
∴ the equation of the tangent at t = 4, i.e. at `(2, 7/2)` is
`y - (7)/(2) = (17)/(4)(x- 2)`
∴ 4y – 14 = 17x – 34
∴ 17x – 4y – 20 = 0
The slope of normal at t = 4
= `(-1)/((dy/dx)_("at" t = 4)`
= `(-1)/((17/4)`
= `-(4)/(17)`
∴ the equation of the normal at t = 4, i.e. at `(2, 7/2)` is
`y - (7)/(2) = -(4)/(17)(x - 2)`
∴ 34y – 119 = – 8x + 16
∴ 8x + 34y – 135 = 0
Hence,, the equations of tangent and normal are
17x – 4y – 20 = 0 and 8x + 34y – 135 = 0 respectively.
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