"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 :"

Question

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

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

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 :

Solution

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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 :

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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 :