Ellipse and vectors

Geometry Level pending

Given that ellipse has the equation: x 2 2020 + y 2 2019 = 1 \dfrac{x^2}{2020}+\dfrac{y^2}{2019}=1 , and line l l always pass through point Q ( 20 , 0 ) Q(\sqrt{20},0) . l l intersects with the y-axis at point P P and the ellipse at point M , N M,N .

If P M = λ 1 M Q , P N = λ 2 N Q \overrightarrow{PM} = \lambda_1 \overrightarrow{MQ}, \overrightarrow{PN} = \lambda_2 \overrightarrow{NQ} , then λ 1 + λ 2 \lambda_1 + \lambda_2 will always be equal to λ \lambda . Find the value of 1000 λ 1000 \lambda .


The answer is -2020.

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1 solution

David Vreken
Apr 27, 2020

Choose line l l to be a horizontal line through Q ( 20 , 0 ) Q(\sqrt{20}, 0) . Then its equation is y = 0 y = 0 , so that it intersects the y y -axis at P ( 0 , 0 ) P(0, 0) the ellipse at M ( 2020 , 0 ) M(-\sqrt{2020}, 0) and N ( 2020 , 0 ) N(-\sqrt{2020}, 0) .

That makes P M = ( 2020 , 0 ) \overrightarrow{PM} = (-\sqrt{2020}, 0) , M Q = ( 2020 + 20 , 0 ) \overrightarrow{MQ} = (\sqrt{2020} + \sqrt{20}, 0) , P N = ( 2020 , 0 ) \overrightarrow{PN} = (\sqrt{2020}, 0) , and N Q = ( 2020 + 20 , 0 ) \overrightarrow{NQ} = (-\sqrt{2020} + \sqrt{20}, 0) , so that λ 1 = P M M Q = 2020 2020 + 20 \lambda_1 = \frac{\overrightarrow{PM}}{\overrightarrow{MQ}} = \frac{-\sqrt{2020}}{\sqrt{2020} + \sqrt{20}} and λ 2 = P N N Q = 2020 2020 + 20 \lambda_2 = \frac{\overrightarrow{PN}}{\overrightarrow{NQ}} = \frac{\sqrt{2020}}{-\sqrt{2020} + \sqrt{20}} , which means λ = λ 1 + λ 2 = 2020 2020 + 20 + 2020 2020 + 20 = 2020 1000 \lambda = \lambda_1 + \lambda_2 = \frac{-\sqrt{2020}}{\sqrt{2020} + \sqrt{20}} + \frac{\sqrt{2020}}{-\sqrt{2020} + \sqrt{20}} = \frac{2020}{1000} .

Therefore, 1000 λ = 1000 2020 1000 = 2020 1000\lambda = 1000 \cdot \frac{2020}{1000} = \boxed{2020} .

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