1.

The line `2y=3x+12` cuts the parabola `4y=3x^(2)`. What is the equation of the hyperbola having rectum and eccentrieity 8 and `3/sqrt5` respectivly ?A. `x^(2)/25+y^(2)/20=1`B. `x^(2)/40+y^(2)/20=1`C. `x^(2)/40+y^(2)/30=1`D. `x^(2)/30+y^(2)/25=1`

Answer» Correct Answer - A
Let the equation of hyperbola be `x^(2)/a^(2)-y^(2)/b^(2)=1`
`"Latus rectum = 8 ="(2b^(2))/arArrb^(2)=4a" …..(i)"`
Also, `b^(2)=a^(2)(e^(2)-1)" [From (i)]"`
`rArr" "4a=a^(2)[(3/sqrt5)^(2)-1]`
`rArr" "a=5&b^(2)=20`
`:." Equation is "x^(2)/25-y^(2)/20=1`


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