This is the question. A parabola $y=ax^2 + bx + x$ has vertex $A(2,1)$ and passes through $B(1,0)$. Find the equation of lines passing through $(0,4)$ that are tangent to the parabola.
Using a system of equations, I've already found the equation of the parabola to be $y=-x^2 + 4x -3$. However, since the point they give is not on the parabola, I have no clue how to find the pair of tangent lines. How should I do it?
On
First we know tangent slope at general point ${(x_1,y_1)}$ on the curve is $$y'=(-2x+4)|_{(x_1,y_1)}$$ which is $y'=-2x_1+4$ now assume that $y'$ is $m$ and find $x_1$ in tems of $m$ and later on you can find $y_1$ in terms of $m$ by the given equation of curve.Now we know $$y-y_1=m(x-x_1)$$ and we know $y_1$ and $x_1$ in terms of $m$.Now convert $y-y_1=m(x-x_1)$ in terms of $m,x,y$ and substitute the point $(0,4)$ in the equation which you got only in terms of $m,x,y$ and solve a quadratic in $m$ which will give you equation of tangents passing through point $(0,4)$.Hope that helps!
On
From the vertex $(2,1)$, the parabola takes the form $y-1=a(x-2)^2$. Plug in the point $(1,0)$ to get $$y= -(x-2)^2+1$$
Let $(p,q)$ be the tangent point and match the slope to get
$$y’= -2(p-2)= \frac{q-4}{p-0}$$
Solve to get $p=\pm \sqrt7$ and the corresponding slopes $\pm 2\sqrt7+4$. Then, use the point-slope formula to get the equations of the tangent lines
$$ y-4=(\pm2\sqrt7+4)x$$
On
The parabola is $$y=-x^2+4x-3~~~~~(1)$$ Let tangent to it from the point $(0,4)$ be $y-4=mx ~~~(2)$, putting this in (1), we get $$mx+4=-x^2+4x-3 \implies x^2+(m-4)x+7=0$$ For tangency demand $B^2=4AC$ in this quadratic, to get $$(m-4)^2=28 \implies m=4\pm 2\sqrt{7}.$$ So the Eq. of tangents is $$y=(4\pm 2 \sqrt{7}) x+4$$
You can do this by taking the derivative of your function as follows:
$y'=-(x^2)'+(4x)'-3'=-2x+4$
Hence the function of the tangent is $-2x+4$.