Please solve this question Ans : Option 1 34 sq units

Question

Aditya Gupta · Answer

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 · Answer

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.

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Please solve this question Ans : Option 1 34 sq units

Please solve this question
Ans : Option 1
34 sq units

2 Answers

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