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# Please solve this question Ans : Option 1              34 sq units

2075 Points
one year ago
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
kindly approve :)
Ankush Bhowmik
13 Points
one year ago
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
$\int_{1}^{2}(4x-4)dx + \int_{2}^{18}(-1/4)(x-18)dx$
Which on solving we get the area of the bounded region as 34 sq. units i.e. option a.