Can someone please help with this math question?

(2a)^2 + a^2 = 80

2ax + ay = 80

Solutions:

a = 4, y = -2 (x - 10)

Hint:

P(2a,a) --->(2a)^2+a^2=80

Tangent QR is 2ax+ay=80

area of the triangle is (1/2)(OQ)(OP)

Perhaps you want it in terms of a ? but you do not ask for that.

angle OPB + angle BPQ = 90º

angle POB + angle OPB = 90º

therefore angle POB = angle BPQ

therefore angle OPB = angle BQP

therefore triangles OBP and BPQ are similar

therefore OB = 2a, BP = a, BQ = a/2

OQ = 2a + (a/2) = (5/2)a

using the same logic RC = 2•2a

and RO = a + 4a = 5a

and area = (1/2)(5/2)(a)(5a) = (5/4)(5)a² = (25/4)a²

You can probably get the area in real numbers... as you know that

OP = √80

therefore a² + (2a)² = 80

5a² = 80

a² = 16

a = 4

area = (25/4)a² = (25/4)16 = 100

Ok, so first of all we need to find out what a is. x = 2a, y = a

(2a)^2 + (a^2) = 80

5a^2 = 80

a = 4

P = (8,4)

So now we need to know the angle of tangent. So we will need to differentiate.

d/dx (x^2 + y^2) = d/dx 80

d/dx (x^2 + y^2) = 0

2x + (2y(dy/dx)) = 0

dy/dx = -2x/2y = -x/y

So we sub in our point (8,4) to get dy/dx = -8/4 = -2

Next we want to know where Q and R are. We know they are along the tangent so we do the following

Q = (8+n,4-(2n)) where n is an unknown

8 + n = 0 so n = -8

y = 20.

Q = (0,20)

R = (8+m,4-(2m))

4-(2m) = 0. So m = 2

8+m = 10

R = (10,0)

We are almost there! So we have a triangle between the points (0,0),(10,0) and (0,20)

The total area is 10*20/2 = 100

Circle: x² + y² = 80

P(2a,a) is a point on the circle, a > 0

4a² + a² = 80

5a² = 80

a² = 16

a = 4 > 0

P(2a,a) = P(8, 4)

Slope of line tangent to circle at P(8, 4)

2x + 2yy' = 0

y' = -x/y = -(8/4) = -2

Tangent Line:

Point-Slope Form: y - 4 = -2(x-8)

Standard Form: 2x + y = 20

Intercept Form: x/10 + y/20 = 1

Tangent Line Intercepts:

y-intercept R: (0, 20)

x-intercept Q: (10, 0)

Area of Triangle OQR:

Base: 10

Height: 20

Area = (1/2)(10)(20) = 100 sq units

Ans: 100 sq units

Okay so (2a)^2 + a^2 = 80 agree?

5a^2 = 80

a = 4 (positive solution in 1st quadrant, but does not affect the final answer)

The line going from origin to point of tangency (radius) has slope 1/2

But the radius is perpendicular to the tangent.

So the tangent line has slope -2.

What line has slope -2 and passes through the point (8, 4)?

y - 4 = -2(x-8)

Find the x- and y-intercepts, then multiply them with 2 and divide (because legs of right triangle)