A hyperbola passes through the point P({2},{3}) and has foci at (± 2,0). Then the tangent to this hyperbola at P also passes through the point : [Main 2017]

Correct Answer is : (32,23)(3{2},2{3})(32,23)

Correct Answer is : (32,23)(3{2},2{3})(32,23)

Test Index

Hyperbola

Section: Mathematics

Question : 91 of 98

Marks: +1, -0

P(\sqrt{2},\sqrt{3})

(\pm 2,0)

Solution:

Equation of hyperbola is

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

foci is

(\pm 2,0)

hence

ae = 2, \Rightarrow a^2 e^2 = 4

b^2 = a^2 (e^2 - 1)

\therefore a^2 + b^2 = 4

......(1)

Hyperbola passes through

(\sqrt{2},\sqrt{3})

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

......(2)

On solving (1) and (2)

a^2 = 8

(is rejected) and

a^2 = 1

and

b^2 = 3

\therefore \frac{x^2}{1} - \frac{y^2}{3} = 1

Equation of tangent id

\frac{\sqrt{2}x}{1} - \frac{\sqrt{3}y}{3} = 1

Hence

(2\sqrt{2},3\sqrt{3})

satisfy it

Go to Question:

Prev Question

Next Question