Please solve this question Ans : Option 134 sq units

by methods of calculus we can easily write eqn of tangent as
y + 4= 2x*2
or 4x= y+4. this intersects x axis at A(1, 0). so AP by dist formula is root(17)
similarly, eqn of normal is y – 4= –1/4(x – 2) or 4y + x= 18. this intersects x axis at B(18, 0). so BP by dist formula is 4root(17).

since tangent and normal are perpendicular to each other, the area of the so formed right triangle APB is ½ * AP * BP.

or ½ *root(17)*4root(17)= 2*17

=34

A

By using calculus we get,

Equation of tangent is 4x-y=4

Equation of normal is X+4y=18

Equation of X axis is y=0

Now, drawing the graphs simultaneously we get the bounded region which we can  divideinto two parts as the sum of two definite integrals as area such as follows

Which on solving we get the area of the bounded region as 34 sq. units i.e. option a.