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The foci of a hyperbola coincide with the foci of the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`. If the eccentricity of the hyperbola is `2`, then the equation of the tangent of this hyperbola passing through the point `(4,6)` isA. `2x-y-2=0`B. `3x-2y=0`C. `2x-3y+10=0`D. `x-2y+8=0` |
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Answer» For the ellipse `(x^(2))/(25)+(y^(2))/(9)=1`, we have `a=5`, `b=3`. `:.e=sqrt(1-(9)/(25))=(4)/(5)` So, the coordinates of the foci are `(+-4,0)`. These are also foci of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` whose eccentricity is `2`. `:.ae=4impliesa=2` Now, `b^(2)=a^(2)(e^(2)-1)impliesb^(2)=4(4-1)=12` So, the equation of the hyperbola is `(x^(2))/(4)-(y^(2))/(12)=1`. Clearly , point `(4,6)` lies on it. The equation of tangent to this hyperbola at `(4,6)` is `(4x)/(4)-(6y)/(12)=1` or, `2x-y=2` |
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