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The equation of the tangent to the curve `y=sqrt(9-2x^(2))` at the point where the ordinate & the abscissa are equal isA. `2x+y-sqrt3=0`B. `2x+y-3=0`C. `2x-y-3sqrt(3)=0`D. `2x+y-3sqrt(3)=0` |
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Answer» Correct Answer - D Let the point `(x,y)` on tangent to the curve Given that `x_(1)=y_(1)` `:.x_(1)=sqrt(9-2x_(1)^(2))` `impliesx_(1)^(2)=9-2x_(1)^(2)impliesx_(1)= +- sqrt(3)` Since `ygt0`, therefore the point is `(sqrt(3),sqrt(3))` Also `y=sqrt(9-2x^(2))impliesy^(2)=9-2x^(2)` differentiate it `2y.(dy)/(dx)=-4ximplies(dy)/(dx)=(-2x)/y` `:.((dy)/(dx))_(((sqrt(3),sqrt(3)))=-2` So, the equation of tangents is `(y-sqrt(3))=-2(x-sqrt(3))` `implies2x+y-3sqrt(3)=0` |
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