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Add maths differentiation question?
Help show me the steps for this question please.
The tangent to a curve y = x(sq) - 4 at the point where x = a intersects the x-axis and the y-axis at points P and Q respectively.
Show that the area of triangle OPQ, where O is the origin is
[a(sq) + 4 ](sq) / 4a.
Thanks
3 Answers
- Anonymous1 decade agoFavorite Answer
y=x^2 - 4
Finding the y co-ordinate when x=a on the cuve, substituting a for x in the equation of the curve.
y=a^2 - 4
Differentiating w.r.t x
dy/dx = 2x
When x=a, dy/dx is equal to
dy/dx = 2a
Means that the gradient of the tangent at the point (a,a^2 - 4) on the curve is 2a.
Now finding the equation of the tangent at (a, a^2 - 4)
y = mx + c
where m = gradient of the line
x and y are respective co-ordinates of any point on the line
Now m = gradient = 2a
and we have one point on that line, that is (a, a^2 - 4)
Substituting the values in the equation y = mx +c
a^2 - 4 = 2a^2 + c
a^2 -2a^2 = c
-a^2 - 4 = c
So the equation is,
y = 2ax - a^2 - 4
The line intersects the y axis when x=0 (You can simply take the value of c directly as y-intercept)
y = 2a(0) - a^2 - 4
y = -a^2 - 4
The line intersects the x axis when y=0
0=2ax - a^2 - 4
(a^2 + 4)/2a=x
Length of OQ = a^2 + 4 (It is taken positive because length cannot be negative)
Length of OP=(a^2 + 4)/2a
Now applying the formula for the area of triangle,
Area of triangle OPQ = 1/2 x base x height
Area of triangle OPQ = 1/2 x OQ x OP
Area of triangle OPQ = 1/2 x (a^2 + 4) x (a^2 + 4)/2a
Area of triangle OPQ = (a^2 + 4)^2/4a
Shown.
- 1 decade ago
You will have to find the equation of the tangent, get the coordinates of P & Q - then you can find the area of the triangle OPQ:
curve: y = x^2 - 4
at x=a, y=a^2-4
dy/dx = d/dx(x^2-4) = 2x
at x=a, dy/dx=2a
Eqn of the tangent: y=x*(dy/dx) + c
y = x*(2a) + c
y = 2ax + c
the line passes through (a, a^2-4)
a^2-4 = 2a^2 + c
c = -a^2 - 4
So:
y = 2ax - a^2 - 4
The line intersects the x-axis at ( (a^2+4)/2a, 0 )
the y-axis at (0, -a^2-4)
The lengths of two sides of the triangle OPQ are:
(a^2 + 4) / 2a
a^2 + 4
Area of the triangle:
= 1/2 * (a^2 + 4) / 2a * (a^2 + 4)
= (a^2+4)^2 / 4a
There's the answer
- 1 decade ago
the tangent is passing by x=a, y=a^2-4, and slope= 2a
so, the tangent equation is y= 2ax-a^2-4
OP= value of x when y is null (this would be (a^2+4)/(2a)
OQ= value of y when x is null (easy, it is a^2+4)
Or area= base x hight /2 = OP x OQ / 2 = (a^2+4)^2/(4a)
