Question

Ans this question plzzz.....​

Answer 1

let the point of tangent be A, intersection point if normal to origin be O,and intersection of tangent at negative x-axis be C

slope of normal=2√2/-1=-2√2

so ,let equation of normal be be 2√2x+y+c=0

slope of tangent=-1/slope of normal=1/2√2

x-2√2y+c'=0=>equation of normal

since Norml pass through (0,0) so,put this value of x and y in normal equation

c=0

so equation of normal=2√2x+y=0

y=-2√2x

also, tangent pass through (-1,2√2)put this value of x and y in the tangent equation

-1-8+c=0

c=9

so equation of tangent=x-2√2y+9=0

2√2y=x+9

y=x/2√2 +9/2√2

y=√2x/4+9√2/4

putting value of y=9,we will get the coordinate where,the tangent,,meets the negative x-axis,i.e the coordinate of third vertex

0=√2x/4+9√2/4

x=-9

so coordinates of third vertex=(-9,0)

now,we have to find the area

split the trangle in two part

A={integration of y=√2x/4 +9√2/4,,from limit y=-9 to -1}+ { integration of x=-2√2x,from x=-1 to 0}

{to make it clear,see the attachment}

A={√2x^2/8 +9√2x/4}( limit from -1 to -9)+{-2√2x^2/2}(limit from -1 to 0}

putting tge limits we get

A={-17√2/8+81√2/8}+√2

A=-17√2/8 +81√2/8 +√2

A=64√2/8 +√2

A=72√2/8

A=9√2

so ,area of traingle will be 9√2(unit)^2