A line parallel to the straight line 2x-y=0 is tangent to the hyperbola x^24-y^22=1 at the…

Question

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

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

Solutions · 1