1.

Find the equation of the tangent of the tangent to the curve 2x^2+3y^2=3 at the point(3,4).(a) x+2y=11(b) x-2y=11(c) -x+2y=11(d) x-2y=-11The question was asked during a job interview.I'd like to ask this question from Derivatives Application in section Application of Derivatives of Mathematics – Class 12

Answer»

The correct option is (a) X+2y=11

The best explanation: Differentiating 2X^2+3y^2=3 w.r.t x, we get

4x+6y \(\frac{dy}{dx}\)=0

\(\frac{dy}{dx} = -\frac{2x}{3y}\)

\(\frac{dy}{dx}\)](3,4)=-\(\frac{2(3)}{3(4)}=-\frac{1}{2}\)

Therefore, the EQUATION of the tangent at (3,4) is

y-y0=m(x-x0)

y-4=-\(\frac{1}{2}\) (x-3)

2(y-4)=-x+3

2y-8=-x+3

x+2y=11


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