"A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola ${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$. Then $x_1^2 + 5y_1^2$ is equal to :"

Question

Accepted Answer

"Tangent of hyperbola ${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$ at the point (x₁, y₁) is

${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$ which is parallel to 2x – y = 0

$\therefore$ Slope of tangent ${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$ = Slope of 2x – y = 0

$\Rightarrow$ ${{{x_1}} \over {2{y_1}}}$ = 2

$\Rightarrow$ x₁ = 4y₁ ....(1)

(x₁, y₁) lies on hyperbola

$\therefore$ ${{x_1^2} \over 4} - {{y_1^2} \over 2} = 1$ ....(2)

From (1) & (2)

${{{{\left( {4{y_1}} \right)}^2}} \over 4} - {{y_1^2} \over 2} = 1$

$\Rightarrow$ $4y_1^2 - {{y_1^2} \over 2} = 1$

$\Rightarrow$ $7y_1^2 = 2$

$\Rightarrow$ $y_1^2 = {2 \over 7}$

Now $x_1^2 + 5y_1^2$

= ${\left( {4{y_1}} \right)^2} + {\left( {5{y_1}} \right)^2}$

= 21$y_1^2$

= 21$\times {2 \over 7}$

= 6"

A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola
${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$. Then $x_1^2 + 5y_1^2$ is equal to :

Solution

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A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola ${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$. Then $x_1^2 + 5y_1^2$ is equal to :

Solution

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More Conic Sections questions

Drill 25 more like these. Every day. Free.

A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola
${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$ at the point $\left( {{x_1},{y_1}} \right)$. Then $x_1^2 + 5y_1^2$ is equal to :