Minimize if you can!
Tangents are drawn from a point on line x − y + 3 = 0 on the curve y 2 = 6 x . For some point P on the on the line, the area of triange formed by tangents and the chord of contact is minimum.
Let this area be △ . Find 9 4 △ .
Try generalize the expression for area of such a traingle for any point P ( x 0 , y 0 ) lying outside the parabola y 2 = 4 a x .
The answer is 4.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Let the general point on the line be: h , k
Chord of contact of parabola: y k = 2 a ( x + h )
Perpendicular distance of h , k from COC(height): k 2 + 4 a 2 k 2 − 4 a h
Length of COC(base): a ( k 2 + 4 a 2 ) × ( k 2 − 4 a h )
Voila!! Now just apply Δ = . 5 × b a s e × h e i g h t and differentiate to get minima.
Due to lack of time, I couldn't completely elucidate the solution.
Let the parabola stand up, y=x^2/6, line y=x-3. It is easier to choose two points on the parabola, (3a, 3a^2/2) and (3c, 3c^2/2). The two tangent line intersect at K= (3[a+c]/2,3ac/2). This can be obtained without calculus, using the law of reflection of a parabola. Then bind two parameters a and c by the condition that K is on line y=x-3: yielding ac=(a+c)-2. We write this as p=s-2, where d is sum, p is product. Let the vertical distance between the midpoint of cord and K be h, which is 3ac/2-3(a^2+c^2)/4=3(a-c)^2. Area=h[3(c-a)/2]=9(c-a)^3/8. To minimize area is to minimize (c-a)^2=s^2-4p=s^2-4s+8. The min=4 at s=2. Plug back into 4Area/9 to obtain 4. Note that in the minimization no calculus is needed, only the vertex of a parabola.
Let the parabola stand up, y=x^2/6, line y=x-3. It is easier to choose two points on the parabola, (3a, 3a^2/2) and (3c, 3c^2/2). The two tangent line intersect at K= (3[a+c]/2,3ac/2). This can be obtained without calculus, using the law of reflection of a parabola. Then bind two parameters a and c by the condition that K is on line y=x-3: yielding ac=(a+c)-2. We write this as p=s-2, where A is sum, p is product. Let the vertical distance between the midpoint of cord and K be h, which is 3ac/2-3(a^2+c^2)/4=3(a-c)^2. Area=h[3(c-a)/2]=9(c-a)^3/8. To minimize area is to minimize (c-a)^2=s^2-4p=s^2-4s+8. The min=4 at s=2. Plug back into 4Area/9 to obtain 4. Note that in the minimization no calculus is needed, only the vertex of a parabola.
The generalised expression for area of triangle formed by pair of tangents from any external point (x1,y1) and the corresponding chord of contact is : [(S1)^(3/2) divided by 2a ] where S1=(y1^2 -4ax1).