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Equation of the tangent to the curve `2x^(2)+3y^(2)-5=0` at (1, 1) isA. `2x-3y-5=0`B. `2x+3y-5=0`C. `2x+3y+5=0`D. `3x+2y+5=0` |
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Answer» Correct Answer - B Equation of the curve is `2x^(2)+3y^(2)-5=0` Differentiating w. r.t.x, we get `4x+6y.(dy)/(dx)=0` `:.(dy)/(dx)=(-4x)/(6y)=(-2x)/(3y)` `:.` Slope of the tangent `=(dy)/(dx)=(-2x)/(3y)` At point `(1,1)((dy)/(dx))_(((1,1)))=(-2)/(3)` `:.` Equation of tangent at (1,1), `y-y_(1)=(dy)/(dx)(x-x_(1))` `y-1=(-2)/(3)(x-1)` `:.3y-3=-2x+2` `2x+3y-5=0` |
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