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Featured 5 months ago

Length of chord

As area of a circle is given by

As shown in the figure, the angle

Featured 4 months ago

Equation of circle is

Let the equation of circle be

As it passes through

Putting

Putting

Now putting

Hence, equation of circle is

Slope of the radius joining

and equation of tangent at **note that it passes through #(0,0)# and hence, this cannot be a tangent, but is a secant** ) as mentioned.

As is seen from figure below, the circle passes through the three points, but

graph{(y-x+3)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

graph{(y+x)(x^2+y^2-3x-3y)(4x-5y)=0 [-3.63, 6.37, -0.96, 4.04]}

Featured 3 months ago

The point of intersection is

The *orthocenter* is where the altitudes intersect in a triangle. Technically, it's where all three altitudes meet. But we only need to find two of them - and it doesn't matter which two.

I've included two pictures - the first one is of the triangle (which is *barely* a triangle) and the other is a picture of the triangle, zoomed out, so we can see where the three altitudes intersect (so far outside the triangle!

So.... Let's find two of the altitudes. We'll need to know two things to find one altitude - one of the vertices, and the slope of the line that the altitude belongs to (we can just use the "rise over run" method for that.

Arbitrarily, let's start with the point

Sweet. Now let's do the same for a different altitude. Again, it doesn't matter which one. Let's pick the point

Sweet.

Now we just need to find the point of intersection of those two altitudes. We can do that by setting them equal to each other.

So the x-coordinate is

So it looks like the point of intersection is

Featured 1 month ago

The given equation of ellipse

Its standard form is

#color(blue)(x^2/a^2+y^2/b^2=1)#

which has its center at origin

Parametrically any point on this ellipse is represented by

The equation of any tagent to this ellipse can be written as

Again by the given condition base of the triangle circumscribing the ellipse is parallel to X-axis.

The tangent for

Now if we put

Hence

It is obvious from the symmetry of the figure that ordinate of the point of intersection of the tangent with y-axis i.e.

and then this ordinate

#OA= b/(sintheta)#

So height of the triangle

#=(b(1+sintheta))/sintheta#

Hence area

Now imposing the condition of minimization i.e.

So

**multiplying bothsides by**

Hence

Inserting the value

Featured 2 weeks ago

Given

and

Now

So

and

Now

slope of straight line [1]

slope of straight line [2]

slope of straight line [3]

**Angle between line [1] and [2]**

**Angle between line [1] and [3]**

So the two angles of the triangle are same and each equals to

Hence the triangle is an equilateral triangle.

Featured 3 weeks ago

shaded area

See the figure above.

Green area

As

Yellow area = area of sector

Green area =

Hence, the shaded area

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