The coordinates of points at each of which the tangents to the curve $y=x^3-3x^2-7x+6$ cut off on the negative semi axis $OX$ a line segment half that on the positive semi axis $OY$ is/are given by
$(A)(-1,9)$
$(B)(3,-15)$
$(C)(1,-3)$
$(D)$none
Let the point of tangency be $(x_1,y_1)$
Then the equation of tangent is $\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$
I am stuck here.
Best Answer
Let the point of tangency be $(x_1,y_1)$
Then the equation of tangent is $\displaystyle\frac{y-y_1}{x-x_1}=3x_1^2-6x_1-7$.
When $y=0$, $\displaystyle x=x_1+\frac{-y_1}{3x_1^2-6x_1-7}$.
When $x=0$, $\displaystyle y=y_1-x_1(3x_1^2-6x_1-7)$.
\begin{align} -2\left(x_1+\frac{-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\ -2\left(\frac{x_1(3x_1^2-6x_1-7)-y_1}{3x_1^2-6x_1-7}\right)&=y_1-x_1(3x_1^2-6x_1-7)\\ 2&=3x_1^2-6x_1-7\\ 3x_1^2-6x_1-9&=0\\ x_1&=-1\quad\text{or}\quad3 \end{align}
When $x_1=-1$, $y_1=(-1)^3-3(-1)^2-7(-1)+6=9$ and the $y$-intercept is
$$9-(-1)[3(-1)^2-6(-1)-7]=11>0$$
When $x_1=3$, $y_1=(3)^3-3(3)^2-7(3)+6=-15$ and the $y$-intercept is
$$-15-(3)[3(3)^2-6(3)-7]=-21<0$$
So, the point of tangency is $(-1,9)$.