A hyperbola passes through the point P(√2,√3) and has foci at (2,0).

Question

A hyperbola passes through the point $P(\sqrt2,\sqrt3)$ and has foci at $(\pm2,0)$. Then the tangent to this hyperbola at $P$ also passes through the point:

Accepted Answer

Equation of hyperbola is $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$

foci is (\pm 2, 0)

$$ \Rightarrow $$ ae = 2

$$ \Rightarrow $$ a2e2 = 4

Since b2 = a2 (e2 -- 1)

b2 = a2 e2 -- a2

$$ \therefore $$ a2 + b2 = 4 .....(1)

Also Hyperbola passes through $$\left( {\sqrt 2 ,\sqrt 3 } \right)$$

$$ \therefore $$ $${2 \over {{a^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ $${2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ (b2 -- 3) (b2 + 4) = 0

$$ \therefore $$ b2 = 3 or b2 = -4

For b2 = 3

$$ \Rightarrow $$ a2 = 1

$${{{x^2}} \over 1} - {{{y^2}} \over 3} = 1$$

Equation of tangent is $${{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1$$

It satisfy point $$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$$.

Solutions · 1