### JEE Main Solved Paper 2017 Question 10

##### Question: A hyperbola passes through the point $ \text{P(}\sqrt{2}\text{,}\sqrt{3}\text{)} $ and has foci at $ (\pm ,2,0). $ Then the tangent to this hyperbola at P is also passes through the point: [JEE Main Solved Paper-2017]

#### Options:

A) $ ( -\sqrt{2},-\sqrt{3} ) $

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

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

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

## Show Answer

#### Answer:

Correct Answer: C

#### 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 $ ?. Hyperbola passes through $ ( \sqrt{2},\sqrt{3} ) $

$ \therefore $ $ \frac{2}{a^{2}}-\frac{3}{b^{2}}=1 $ ?. On solving [a] and [b] $ a^{2}=8 $ (is rejected) and $ a^{2}=1 $ and $ b^{2}=3 $

$ \therefore $ $ \frac{x^{2}}{1}-\frac{y^{3}}{3}=1 $ Equation of tangent is $ \frac{\sqrt{2x}}{1}-\frac{\sqrt{3y}}{3}=1 $ Hence $ ( 2\sqrt{2},3\sqrt{3} ) $ satisfy it.