A hyperbola passes through the point `(sqrt2,sqrt3)` and has foci at `(+-2,0)`. Then the tangent to this hyperbola at P also passes through the point:
A
`(-sqrt2,-sqrt3)`
B
`(3sqrt2,2sqrt3)`
C
`(2sqrt2,3sqrt3)`
D
`(sqrt3,sqrt2)`
Text Solution
Verified by Experts
The correct Answer is:
C
Equation of hyperbola is `(X^(2))/(a^(2))-(y^(2))/(b^(2))=1`. Foci are `(pm2,0)`. So , ae = 2. Now, `b^(2)=a^(2)(e^(2)-1)` `therefore" "a^(2)+b^(2)=4" (1)"` Given hyperbola passes through `(sqrt2,sqrt3).` `therefore" "(2)/(a^(2))-(3)/(b^(2))=1" (2)"` One solving (1) and (2), we get `a^(2)=1 and b^(2)=3` So, equation of the hyperbola is `(x^(2))/(1)-(y^(2))/(3)=1`. Hence, equation of tangent at `P(sqrt2,sqrt3)" is "(sqrt2x)/(1)-(sqrt3y)/(3)=1.` which passes through the point `(2sqrt2,3sqrt3)`.
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