Question

Find the equation of the line tangent to the graph of x(t)=t²+1 and y(t)=

Answer 1

Answer:

The equation of the straight line

  x - 24y + 38 =0

Step-by-step explanation:

Step(i):-

Given that x(t) = t²+1 ..(i)

        and y(t) = √1+t  ..(ii)

Differentiating equation(i) with respective to 't'

$\frac{dx}{dt} = 2t$

Differentiating equation(ii) with respective to 't'

$\frac{dy}{dt} = \frac{1}{2\sqrt{1+t} }$

Step(ii):-

The slope of the tangent

     $m = \frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} } = \frac{\frac{1}{2\sqrt{1+t} } }{2t}$

    $m =( \frac{dy}{dx} )_{t=3} = \frac{1}{4(3)\sqrt{1+3} }$

    $m = \frac{1}{24}$

Step(iii):-

Point   x = t²+1 = 3²+1 = 10

          y = √1+t =√1+3 = √4 =2

The point on the tangent line is ( 10 ,2)

The equation of the straight line

      $y - y_{1} = m( x-x_{1} )$

     $y - 2 = \frac{1}{24} ( x-10 )$

   24 (y-2) = x-10

   24y - 48 = x-10

    x - 24 y -10 +48 =0

   x - 24y + 38 =0

Final answer:-

The equation of the straight line

  x - 24y + 38 =0