To find the length of the shortest path that begins at $(-1,1)$, touches the x-axis and then ends at a point on the parabola $(x-y)^2 = 2(x + y −4)$:

Question

This is a question of a parabola but to solve it I need the coordinates of the vertex and focus which at present I'm unable to deduce in this form. I have learnt some standard form such as $ y^2=4ax$ and other such forms. Can anyone explain how to locate the vertex, focus and also how to get the equation of the directrix.

Actual question : The length of the shortest path that begins at the point $(-1, 1)$, touches the x-axis and then ends at a point on the parabola $(x-y)^2 = 2(x + y −4)$, is?

Answer 1

Parabola equation is

$ (x - y)^2 = 2 (x + y - 4) $

Reflect the point $(-1, 1)$ about the $x$ axis. You will get $(-1, -1)$.

Now we need to connect $(-1,-1)$ to the parabola by a line segment that is perpendicular to its curve.

Compute the slope of the tangent to the parabola by differnetiating with respect to $x$. This will give

$ 2 (x - y) (1 - y' ) = 2 (1 + y') $

Hence

$ y' = \dfrac{x - y - 1}{x - y + 1} $

Now the line segment connecting $(x,y)$ to $(-1, -1)$ has a slope of

$ m = \dfrac{y+1}{x+1} $

The product of $m $ and $y'$ has to be equal to $(-1) $.

$ -1 = \dfrac{ (y+1) (x - y - 1)}{ (x+1)(x - y + 1) } $

i.e.

$ - (x + 1)(x - y + 1) = (y + 1)(x - y - 1) $

Expand,

$ -( x^2 - x y + x + x - y + 1) = x y - y^2 - y + x - y - 1 $

This simplifies to

$ - x^2 + y^2 - 3 x + 3 y = 0 $

This factors into

$ (y - x) ( 3 + y + x) = 0 $

This implies either $ y = x$ or $ x + y = -3 $

If $ x + y = -3 $, the plug this into the parabola equation

$ (x - y)^2 = 2 ( -3 - 4 ) $

which has no real solutions.

The other solution is $ x = y $ and this gives

$ 0 = 2 (2 y - 4) $

So $ y = x = 2 $

Connect $(2, 2)$ to $(-1, -1)$ gives the minimum distance as $3 \sqrt{2}$

Answer 2

I can help you with finding vertex and focus of the parabola you can rewrite the the equation $y^2=4ax$ as : $$(\text{distance from x axis})^2=4a(\text{distance from y axis})$$ in this case x axis is the axis of parabola and y axis is tangent at vertex.

we know that distance of a point $(x_1,y_1)$ from a line $y=mx+c$ is $$|\frac{mx_1-y_1+c}{\sqrt{m^1+1}}|$$

here judging from the equation tangent at vertex is $x+y-4=0$ and axis of parabola is $x=y$

so equation of parabola should be of form : $$\left(\frac{x-y}{\sqrt 2}\right)^2=4a|\frac{x+y-4}{\sqrt2}|$$ $$(x-y)^2=4a\sqrt 2 (x+y-4)$$ from the given equation we know that $4a\sqrt2=2$ so $a=\frac{\sqrt2}{4}$

by solving two lines we know that vertex is (2,2) and by using some geometry and value of a we can calculate that focus is $\left(\frac{9}{4},\frac{9}{4}\right)$