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If the tangent at the point \mathrm{\left(x_{1}, y_{1}\right)} on the curve \mathrm{y=x^{3}+3 x^{2}+5} passes through the origin, then \mathrm{\left(x_{1}, y_{1}\right)}  does NOT lie on the curve :

Option: 1

\mathrm{x^{2}+\frac{y^{2}}{81}=2}


Option: 2

\mathrm{\frac{y^{2}}{9}-x^{2}=8}


Option: 3

\mathrm{y=4 x^{2}+5}


Option: 4

\mathrm{\frac{x}{3}-y^{2}=2}


Answers (1)

best_answer

\mathrm{y_{1}= x_{1}\, ^{3}+3\, x_{1}\, ^{2}+5 \quad \cdots(i)}

\mathrm{\frac{dy}{dx}= 3\, x_{1}\, ^{2}+6\, x_{1}}
\mathrm{y-y_{1}= \left ( 3\, x_{1}\, ^{2}+6\, x_{1} \right )\left ( x-x_{1} \right )}

Passing through origin :
\mathrm{y_{1}= \left ( 3\, x_{1}\, ^{2}+6\, x_{1} \right )\left ( x_{1} \right )\quad \cdots(ii)}

From  equation (i) & (ii),
\mathrm{ x_{1}\, ^{3}+3\, x_{1}\, ^{2}+5= 3\, x_{1}\, ^{3}+6\, x_{1}\, ^{2}}
\mathrm{ 2\, x_{1}\, ^{3}+3\, x_{1}\, ^{2}-5= 0}
\mathrm{ \left ( x_{1}-1 \right )\left ( 2\, x_{1}\, ^{2}+5\, x_{1} +5\right )= 0}
\mathrm{ x_{1}= 1}

\mathrm{ \therefore \: \: point:\left ( 1,9 \right )}
The correct answer is option (D)

Posted by

sudhir.kumar

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