59. Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2 , 2at).

The given of the parabola is y2=4axslope of tangent is given byddxy2=ddx4ax→2ydydx=4a⇒dydx=2ay.dydx|(at2,2at)=2a2at=1tso, slope of normalso, slope of normal =−1(1t)=−tHence eqn of tangent at point (at2,2at) isy−2at=1t(x−at2)→ty−2at2=x−at2→x−ty−at2+2at2=0x−ty+at2=0And eqn of normal at point (at2,2at) is(y−2at)=(−t)(x−at2)→y−2at=−tx+at3⇒tx+y−2at+at3=0⇒tx+y−at(2+t2)=0

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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

59. Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at² , 2at).

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Vishal Baghel | Contributor-Level 10

4 months ago

The given of the parabola is $y^{2} = 4 a x$

slope of tangent is given by

$\frac{d}{d x} y^{2} = \frac{d}{d x} 4 a x$

$\to 2 y \frac{d y}{d x} = 4 a$

$\Rightarrow \frac{d y}{d x} = \frac{2 a}{y} .$

${\frac{d y}{d x} |}_{(a t^{2}, 2 a t)} = \frac{2 a}{2 a t} = \frac{1}{t}$

so, slope of normal

so, slope of normal $= \frac{- 1}{(\frac{1}{t})} = - t$

Hence eqⁿ of tangent at point $(a t^{2}, 2 a t)$ is

$y - 2 a t = \frac{1}{t} (x - a t^{2})$

$\to t y - 2 a t^{2} = x - a t^{2}$

$\to x - t y - a t^{2} + 2 a t^{2} = 0$

$x - t y + a t^{2} = 0$

And eqⁿ of normal at point $(a t^{2}, 2 a t)$ is

$(y - 2 a t) = (- t) (x - a t^{2})$

$\to y - 2 a t = - t x + a t^{3}$

$\Rightarrow t x + y - 2 a t + a t^{3} = 0$

$\Rightarrow t x + y - a t (2 + t^{2}) = 0$

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y (x) = ∫? (2t² - 15t + 10)dt
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By truth table

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