Find the equations of tangents to the curve $3x^2 - y^2 = 8$ which pass through the point $(4/3,0)$
I googled it and got this yahoo answers question which has 2 contradicting answers.
I did it the first way. but got m = 1/3 and the points of tangency are ( 2 , 2 ) & ( 2 , - 2) But my book does it the second way but reaches the equation y=0 (slope is 0)
I am absolutely confused
Also, I found a contradiction here:$$x-x_1=m(y-y_1) $$ $$x-x_1=\frac{a}b{} (y-y_1)$$ $$bx-bx_1=ay-ay_1$$ $$y=\frac{bx-bx_1+ay_1}{a}$$ $$y=mx$$ Here slope m becomes b/a instead of a/b. What did i do wrong?
I have exam tmrw. Quick answer would be appreciated. Thank you
The slope of the line passing through $(x,y)$ and $(4/3,0)$ is given by: $$ m_1(x,y) = \frac{y}{x-4/3}. $$ The tangent line passing through $(x,y)$ on your curve is perpendicular to the gradient of your function: $$ Df(x,y) = \left(\begin{array}{c}6x \\ -2y\end{array}\right) $$ so it is parallel to the vector $$ \left(\begin{array}{c}2y \\ 6x\end{array}\right) $$ which has slope $$ m_2(x,y) = \frac{6x}{2y}. $$ It must be $m_1(x,y)=m_2(x,y)$ i.e. one must find solutions of the system: $$ \begin{cases} \frac{y}{x-4/3} = \frac{3x}{y}\\ 3x^2 - y^2 = 8 \end{cases} $$