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Please solve this question Ans : Option 1 34 sq units
by methods of calculus we can easily write eqn of tangent asy + 4= 2x*2or 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=34kindly approve :)
ABy using calculus we get,Equation of tangent is 4x-y=4Equation of normal is X+4y=18Equation of X axis is y=0Now, 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 followsWhich on solving we get the area of the bounded region as 34 sq. units i.e. option a.
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