if the tangents on the ellipse `4x^(2)+y^(2)=8` at the points (1,2) and (a,b) are perpendicular to each other then `a^(2)` is equal to

Options

A. `(128)/(17)`
B. `(64)/(17)`
C. `(4)/(17)`
D. `(2)/(17)`

Answer

D

Explanation

Equation of given ellipse is
`4x^(2) + y^(2) = 8`
` implies (x^(2) )/(2) + y^(2))/(8) =1=? ( x^(2))/((sqrt(2))^(2))+ ( y^(2))/((2sqrt(2))^(2))=1`
Now, equation of tangent to the ellipse `(x^(2))/( a^(2))+(y^(2))/( b^(2))=1 ` at `( x_(1),x_(1))` is `(x x_(1))/( a^(2))+(y y_(1))/(b^(2))=1]`
and equation of another tangent at point ( a,b) is
`4ax+by=8`
since lines (ii) and (iii) are perpendicular to each other .
`therefore (-(2)/(1))xx(-(4a)/(b))=-1`
if lines `a_(1)x+b_(1) y+c_(1)=0 and a_(2) x+ b_(2) y+c_(2) =0`
are per[endicular , then `(-a_(1))/( b_(1))(-(a_(2))/(b_(2))=-1]`
`implies b=8a `
also , the point (a,b) lies on the ellipes (i) so ,
`implies 4a^(2) + b^(2)=8`
`implies 4a^(2)+b^(2)=8` [ from Eq.(iv)]
`implies 4a^(2) + 64 a^(2)=8`
`implies 687 a^(2)= 8 implies a^(2)=(8) /(68)`
`implies a^(2)=(2)/(17)`