Find the equation of the tangent line at $y^2=6x-3$ that is perpendicular to the line $x+3y=7$
My answer : $y=(2\sqrt{30}-9)+3(x-34+6\sqrt{30})$
Graphing the line seems close enough but after more than 10 pages of paper I don't really know what to do anymore. send help.
On
Hint:
The equation of any straight line perpendicular to $x+3y=7$ is $$3x-y=k$$
Now let us find the intersection of this line with $y^2=6x-3$
$$y^2=2(y+k)-3\iff y^2-2y+3-2k=0\ \ \ \ (1)$$ which is a Quadratic Equation $y$ whose roots represent the ordinates of intersection.
For tangency, both roots must coincide.
So, the discriminant of $(1)$ must be $0$
Can you take it from here?
Line has slope $-(1/3)$, recall $-(1/3)m=-1$ gives $m$ , the perpendicular slope.
Hence: $m=3;$
Differentiate $y^2=6x -3$.
$2y\dfrac{dy}{dx} =6;$
Setting $\dfrac{dy}{dx} =3$:
$2y(3)= 6;$ $ y=1.$
$6x-3=1$; $6x=4$; $x= 2/3$.
Tangent line:
$y-1= 3(x-2/3).$