Question

A trapezium is inscribed in the parabola y2 = 4x, such that its diagonal pass through the point (1,0) andeach has length 25/4- If the area of the trapezium be P, then 4P is equal to:(B) 71(C) 80(D) 75(A) 70​

Answer 1

Answer:

4P is equal to 75 square units.

Step-by-step explanation:

Let’s draw a parabola y² = 4x, on the x-y plane. We are given that the trapezium is inscribed inside the parabola so the we will take the point ABCD i.e., the corners of the trapezium such that all the corners touches the parabola.  

Also, let the focus of the trapezium be “S” which is given as (1,0) through which the diagonal passes and is of length 25/4. Shown in the fig. below

Now let the distance AS = C and we know that

1/AS + 1/CS = 1/AC … (i)

AC=25/4 and AS=C  

∴ CS = 25/4 – C and AC = 1…. (ii)

Substituting (ii) in (i), we get

1/C + 1/[(25/4)-C] = 1/1

Or, 25/4 = 25C/4 – C²

Or, 4C² – 25C + 25 = 0

Or, 4C(C-5) – 5 (C-5) = 0

Or, C = 5 or 5/4

Since AC is the focal chord therefore we can write, AS = C = (1+t²)  

When C = 5/4,

C=1+t²

Or, 5/4 = 1 + t²

Or, t = ± ½  

And, when C = 5,

C = 1 + t²

Or, 5 = 1 + t²

Or, t = ±2

Now, we can will determine the coordinates of the trapezium ABCD

Putting the value of a=1 and t=±1/2  

Coordinates of A(at², 2at): A(1*1/4, 2*1*1/2)=A(1/4,1)

Coordinates of D(at², 2at): D(1*1/4, 2*1*-1/2)=D(1/4,-1)

Coordinates of B( a/t² , 2a/t): B(1/(1/4), 2*1/(1/2))=B(4,4)

Coordinates of C( a/t² , 2a/t): C(1/(1/4), 2*1/(-1/2))=C(4,-4)

∴ Distance AD = 1 – (-1) = 2 units

∴ Distance BC = 4 – (-4) = 8 units

∴ Distance between AD & BC = 4 – (1/4) = 15/4 units

According to the question we are given the area of trapezium as P and we have to calculate the value of 4P.

∴ Area of trapezium ABCD, P = ½ *(sum of parallel sides)*(distance between the parallel sides)

          = ½ * (2+8) * 15/4

           = (10 * 15) / (2*4)

           = 75/4 square units

So, 4P = 4 * 75/4 = 75 square units.