Answer
Verified
407.1k+ views
Hint:
Here, we will first draw a figure to show the given condition. We will then find the focal chord and the coordinates of the trapezium. Using this we will find the sides and the height of the trapezium. Then we will substitute these values in the formula of area of trapezium and hence, find the required value of \[4P\].
Formula Used:
Area of trapezium \[ABCD = \dfrac{1}{2} \times \] sum of parallel sides \[ \times \] height
Complete step by step solution:
First we will draw a parabola \[{y^2} = 4x\] which opens toward the positive side of the \[x\] axis.
Now, since, the trapezium is inscribed in this parabola, we will draw a trapezium such that its corners touch the parabola.
Now, let the focus of the trapezium be \[S\left( {1,0} \right)\] through which the diagonals of the trapezium passes and are of the length\[\dfrac{{25}}{4}\].
Now, let \[AS = x\] and, diagonal \[AC = \dfrac{{25}}{4}\]. Now we can write diagonal AC as a sum of AS and SC. So,
\[AS + SC = AC\]…………………………….\[\left( 1 \right)\]
Substituting \[AS = x\] and \[AC = \dfrac{{25}}{4}\] in the above equation, we get
\[ \Rightarrow x + SC = \dfrac{{25}}{4}\]
Subtracting \[x\] from both the sides, we get
\[ \Rightarrow SC = \dfrac{{25}}{4} - x\]
Now taking reciprocal of equation \[\left( 1 \right)\], we can write it as:
\[\dfrac{1}{{AS}} + \dfrac{1}{{SC}} = \dfrac{1}{{AC}}\]……………………….\[\left( 2 \right)\]
Substituting \[SC = \dfrac{{25}}{4} - x\], \[AS = x\] and \[AC = 1\] in equation \[\left( 2 \right)\], we get
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{\left( {\dfrac{{25}}{4} - x} \right)}} = \dfrac{1}{1}\]
Taking LCM on left hand side of the above equation, we get,
\[ \Rightarrow \dfrac{{\left( {\dfrac{{25}}{4} - x + x} \right)}}{{x\left( {\dfrac{{25}}{4} - x} \right)}} = 1\]
On cross multiplication, we get
\[ \Rightarrow \dfrac{{25}}{4} = \dfrac{{25}}{4}x - {x^2}\]
Taking LCM on RHS, we get
\[ \Rightarrow \dfrac{{25}}{4} = \dfrac{{25x - 4{x^2}}}{4}\]
\[ \Rightarrow 25 = 25x - 4{x^2}\]
Rewriting the above equation, we get
\[ \Rightarrow 4{x^2} - 25x + 25 = 0\]
The above equation is a quadratic equation, so we will factorize the equation to get the value of \[x\].
Now, splitting the middle term, we get
\[ \Rightarrow 4{x^2} - 20x - 5x + 25 = 0\]
\[ \Rightarrow 4x\left( {x - 5} \right) - 5\left( {x - 5} \right) = 0\]
Factoring out the common term, we get
\[ \Rightarrow \left( {4x - 5} \right)\left( {x - 5} \right) = 0\]
Applying zero product property, we get
\[ \Rightarrow \left( {4x - 5} \right) = 0\] or \[\left( {x - 5} \right) = 0\]
Hence,
\[x = \dfrac{5}{4}\] or \[x = 5\]
Now, since \[AC\] is the focal chord, hence, we can write,
\[AS = x = \left( {1 + {t^2}} \right)\]………………………..\[\left( 3 \right)\]
Now, substituting \[x = \dfrac{5}{4}\] in the above equation, we get
\[ \Rightarrow \dfrac{5}{4} = \left( {1 + {t^2}} \right)\]
Subtracting 1 from both sides, we get
\[ \Rightarrow \dfrac{5}{4} - 1 = {t^2}\]
\[ \Rightarrow \dfrac{1}{4} = {t^2}\]
Rewriting the above equation, we get
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {t^2}\]
Taking square root on both sides, we get
\[ \Rightarrow t = \pm \dfrac{1}{2}\]
Substituting \[x = 5\] in equation \[\left( 3 \right)\], we get
\[ \Rightarrow 5 = \left( {1 + {t^2}} \right)\]
Subtracting 1 from both sides, we get
\[\begin{array}{l} \Rightarrow {t^2} = 4\\ \Rightarrow {t^2} = {\left( 2 \right)^2}\end{array}\]
Taking square root on both sides, we get
\[ \Rightarrow t = \pm 2\]
Now, we will find the coordinates of the trapezium \[ABCD\]
Substituting \[a = 1\] and \[t = \pm \dfrac{1}{2}\] in the coordinates of the trapezium \[ABCD\], we get
Coordinates of \[A\left( {a{t^2},2at} \right) = \left( {1 \times \dfrac{1}{4},2 \times 1 \times \dfrac{1}{2}} \right) = A\left( {\dfrac{1}{4},1} \right)\],
Coordinates of \[D\left( {a{t^2},2at} \right) = \left( {1 \times \dfrac{1}{4},2 \times 1 \times \dfrac{{ - 1}}{2}} \right) = A\left( {\dfrac{1}{4}, - 1} \right)\]
Coordinates of \[B\left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) = \left( {\dfrac{1}{{\dfrac{1}{4}}},\dfrac{{2 \times 1}}{{\dfrac{1}{2}}}} \right) = \left( {4,4} \right)\]
Coordinates of \[C\left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) = \left( {\dfrac{1}{{\dfrac{1}{4}}},\dfrac{{2 \times 1}}{{\dfrac{{ - 1}}{2}}}} \right) = \left( {4, - 4} \right)\]
Now, by distance formula, \[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \], we get
Distance of \[AD = \sqrt {{{\left( {\dfrac{1}{4} - \dfrac{1}{4}} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} = \sqrt 4 = 2{\rm{ units}}\]
Distance of \[BC = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( { - 4 - 4} \right)}^2}} = \sqrt {64} = 8{\rm{ units}}\]
$\therefore $ Height of the trapezium or Distance between the lines \[AD\] and \[BC = 4 - \dfrac{1}{4} = \dfrac{{15}}{4}{\rm{ units}}\]
Now, according to the question, the area of the trapezium is \[P\]
As we know,
Area of trapezium \[ABCD = \dfrac{1}{2} \times \]sum of parallel sides\[ \times \]height
\[ \Rightarrow P = \dfrac{1}{2}\left( {2 + 8} \right)\left( {\dfrac{{15}}{4}} \right)\]
Adding the terms in the bracket, we get
\[ \Rightarrow P = \dfrac{1}{2} \times 10 \times \left( {\dfrac{{15}}{4}} \right) = 5 \times \dfrac{{15}}{4}\]
Multiplying the terms, we get
\[ \Rightarrow P = \dfrac{{75}}{4}\]
$\therefore 4P=4\times \dfrac{75}{4}=75$ square units.
Therefore, if the area of the trapezium is \[P\], then \[4P\] is equal to 75 square units.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. Whereas, a trapezium is a quadrilateral which has one pair of its opposite sides parallel. Now, in this question, a trapezium is inscribed in a parabola. Hence, we will draw the largest possible trapezium which can be drawn inside a parabola to solve this question.
Here, we will first draw a figure to show the given condition. We will then find the focal chord and the coordinates of the trapezium. Using this we will find the sides and the height of the trapezium. Then we will substitute these values in the formula of area of trapezium and hence, find the required value of \[4P\].
Formula Used:
Area of trapezium \[ABCD = \dfrac{1}{2} \times \] sum of parallel sides \[ \times \] height
Complete step by step solution:
First we will draw a parabola \[{y^2} = 4x\] which opens toward the positive side of the \[x\] axis.
Now, since, the trapezium is inscribed in this parabola, we will draw a trapezium such that its corners touch the parabola.
Now, let the focus of the trapezium be \[S\left( {1,0} \right)\] through which the diagonals of the trapezium passes and are of the length\[\dfrac{{25}}{4}\].
Now, let \[AS = x\] and, diagonal \[AC = \dfrac{{25}}{4}\]. Now we can write diagonal AC as a sum of AS and SC. So,
\[AS + SC = AC\]…………………………….\[\left( 1 \right)\]
Substituting \[AS = x\] and \[AC = \dfrac{{25}}{4}\] in the above equation, we get
\[ \Rightarrow x + SC = \dfrac{{25}}{4}\]
Subtracting \[x\] from both the sides, we get
\[ \Rightarrow SC = \dfrac{{25}}{4} - x\]
Now taking reciprocal of equation \[\left( 1 \right)\], we can write it as:
\[\dfrac{1}{{AS}} + \dfrac{1}{{SC}} = \dfrac{1}{{AC}}\]……………………….\[\left( 2 \right)\]
Substituting \[SC = \dfrac{{25}}{4} - x\], \[AS = x\] and \[AC = 1\] in equation \[\left( 2 \right)\], we get
\[ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{\left( {\dfrac{{25}}{4} - x} \right)}} = \dfrac{1}{1}\]
Taking LCM on left hand side of the above equation, we get,
\[ \Rightarrow \dfrac{{\left( {\dfrac{{25}}{4} - x + x} \right)}}{{x\left( {\dfrac{{25}}{4} - x} \right)}} = 1\]
On cross multiplication, we get
\[ \Rightarrow \dfrac{{25}}{4} = \dfrac{{25}}{4}x - {x^2}\]
Taking LCM on RHS, we get
\[ \Rightarrow \dfrac{{25}}{4} = \dfrac{{25x - 4{x^2}}}{4}\]
\[ \Rightarrow 25 = 25x - 4{x^2}\]
Rewriting the above equation, we get
\[ \Rightarrow 4{x^2} - 25x + 25 = 0\]
The above equation is a quadratic equation, so we will factorize the equation to get the value of \[x\].
Now, splitting the middle term, we get
\[ \Rightarrow 4{x^2} - 20x - 5x + 25 = 0\]
\[ \Rightarrow 4x\left( {x - 5} \right) - 5\left( {x - 5} \right) = 0\]
Factoring out the common term, we get
\[ \Rightarrow \left( {4x - 5} \right)\left( {x - 5} \right) = 0\]
Applying zero product property, we get
\[ \Rightarrow \left( {4x - 5} \right) = 0\] or \[\left( {x - 5} \right) = 0\]
Hence,
\[x = \dfrac{5}{4}\] or \[x = 5\]
Now, since \[AC\] is the focal chord, hence, we can write,
\[AS = x = \left( {1 + {t^2}} \right)\]………………………..\[\left( 3 \right)\]
Now, substituting \[x = \dfrac{5}{4}\] in the above equation, we get
\[ \Rightarrow \dfrac{5}{4} = \left( {1 + {t^2}} \right)\]
Subtracting 1 from both sides, we get
\[ \Rightarrow \dfrac{5}{4} - 1 = {t^2}\]
\[ \Rightarrow \dfrac{1}{4} = {t^2}\]
Rewriting the above equation, we get
\[ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {t^2}\]
Taking square root on both sides, we get
\[ \Rightarrow t = \pm \dfrac{1}{2}\]
Substituting \[x = 5\] in equation \[\left( 3 \right)\], we get
\[ \Rightarrow 5 = \left( {1 + {t^2}} \right)\]
Subtracting 1 from both sides, we get
\[\begin{array}{l} \Rightarrow {t^2} = 4\\ \Rightarrow {t^2} = {\left( 2 \right)^2}\end{array}\]
Taking square root on both sides, we get
\[ \Rightarrow t = \pm 2\]
Now, we will find the coordinates of the trapezium \[ABCD\]
Substituting \[a = 1\] and \[t = \pm \dfrac{1}{2}\] in the coordinates of the trapezium \[ABCD\], we get
Coordinates of \[A\left( {a{t^2},2at} \right) = \left( {1 \times \dfrac{1}{4},2 \times 1 \times \dfrac{1}{2}} \right) = A\left( {\dfrac{1}{4},1} \right)\],
Coordinates of \[D\left( {a{t^2},2at} \right) = \left( {1 \times \dfrac{1}{4},2 \times 1 \times \dfrac{{ - 1}}{2}} \right) = A\left( {\dfrac{1}{4}, - 1} \right)\]
Coordinates of \[B\left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) = \left( {\dfrac{1}{{\dfrac{1}{4}}},\dfrac{{2 \times 1}}{{\dfrac{1}{2}}}} \right) = \left( {4,4} \right)\]
Coordinates of \[C\left( {\dfrac{a}{{{t^2}}},\dfrac{{2a}}{t}} \right) = \left( {\dfrac{1}{{\dfrac{1}{4}}},\dfrac{{2 \times 1}}{{\dfrac{{ - 1}}{2}}}} \right) = \left( {4, - 4} \right)\]
Now, by distance formula, \[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \], we get
Distance of \[AD = \sqrt {{{\left( {\dfrac{1}{4} - \dfrac{1}{4}} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} = \sqrt 4 = 2{\rm{ units}}\]
Distance of \[BC = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( { - 4 - 4} \right)}^2}} = \sqrt {64} = 8{\rm{ units}}\]
$\therefore $ Height of the trapezium or Distance between the lines \[AD\] and \[BC = 4 - \dfrac{1}{4} = \dfrac{{15}}{4}{\rm{ units}}\]
Now, according to the question, the area of the trapezium is \[P\]
As we know,
Area of trapezium \[ABCD = \dfrac{1}{2} \times \]sum of parallel sides\[ \times \]height
\[ \Rightarrow P = \dfrac{1}{2}\left( {2 + 8} \right)\left( {\dfrac{{15}}{4}} \right)\]
Adding the terms in the bracket, we get
\[ \Rightarrow P = \dfrac{1}{2} \times 10 \times \left( {\dfrac{{15}}{4}} \right) = 5 \times \dfrac{{15}}{4}\]
Multiplying the terms, we get
\[ \Rightarrow P = \dfrac{{75}}{4}\]
$\therefore 4P=4\times \dfrac{75}{4}=75$ square units.
Therefore, if the area of the trapezium is \[P\], then \[4P\] is equal to 75 square units.
Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. Whereas, a trapezium is a quadrilateral which has one pair of its opposite sides parallel. Now, in this question, a trapezium is inscribed in a parabola. Hence, we will draw the largest possible trapezium which can be drawn inside a parabola to solve this question.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Which are the Top 10 Largest Countries of the World?
One cusec is equal to how many liters class 8 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
The mountain range which stretches from Gujarat in class 10 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths