If the normal from the point P(h,1) on the ellipse `x^2/6+y^2/3=1` is perpendicular to the line `x+y=8` , then the value of h is

Options

A. 1
B. 2
C. 8
D. 9

Answer

B

Explanation

The slope of the normal to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at the point `(x_(1),y_(1))` is `(a^(2)y_(1))/(b^(2)x_(1))`. So, the slope of the normal to `(x^(2))/(6)+(y^(2))/(3)=1 P(h,1)" is " (6)/(3h)=(2)/(h)`.
It is given that the normal is perpendicular to the line x + y = 8.
`therefore (2)/(h)xx-1rArrh=2`