Let the line y = mx and the ellipse 2x2 + y2 = 1 intersect at a ponit P in the first quadr…

Question

Let the line y = mx and the ellipse 2x2 + y2 = 1
intersect at a ponit P in the first quadrant. If the
normal to this ellipse at P meets the co-ordinate axes at $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$ and (0, $$\beta $$), then $$\beta $$ is equal
to :

Accepted Answer

Let P be (x1
, y1)

Equation of normal at P is $${x \over {2{x_1}}} - {y \over {{y_1}}} = {1 \over 2} - 1$$

It passes through $$\left( { - {1 \over {3\sqrt 2 }},0} \right)$$

$$ \therefore $$ $${{ - 1} \over {6\sqrt 2 {x_1}}} =  - {1 \over 2}$$

$$ \Rightarrow $$ x1 = $${1 \over {3\sqrt 2 }}$$

Also using 2$x_1^2$ + $y_1^2$ = 1

$$ \Rightarrow $$ $y_1^2$ = 1 - $${2 \over {18}}$$

$$ \Rightarrow $$ y1 = $${{2\sqrt 2 } \over 3}$$

(0, $$\beta $$) is on the normal.

So 0 - $${\beta  \over {{{2\sqrt 2 } \over 3}}}$$ = $$ - {1 \over 2}$$

$$ \Rightarrow $$ $$\beta $$ = $${{\sqrt 2 } \over 3}$$

Solutions · 1