Let ABCD be a trapezium whose vertices lie on the parabola y^2=4 x . Let the sides AD and BC of the trapezium be parallel to y -axis. If the diagonal AC is of length \frac254 and it passes through the point (1,0) , then the area of A B C D is

Parabola PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q37 with Solution

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Correct answer: D

Parametrize points on the parabola

For the parabola $y^2=4x$ , a general point can be written as $P(t)=(t^2,2t).$

Since $AD$ and $BC$ are parallel to the $y$ -axis, points on each of these sides must have the same $x$ -coordinate.

So take:

$A=(a^2,2a)$ and $D=(a^2,-2a)$ ,
$B=(b^2,2b)$ and $C=(b^2,-2b)$ .

Thus $AD$ and $BC$ are vertical chords.

Use the fact that diagonal $AC$ passes through $(1,0)$

Points $A$ and $C$ are:

\qquad C=(b^2,-2b).$$ Since $(1,0)$ lies on line $AC$, the slope from $A$ to $(1,0)$ equals the slope from $(1,0)$ to $C$. Using section/collinearity form: Equation of line through $A$ and $C$: $$\frac{y-2a}{-2b-2a}=\frac{x-a^2}{b^2-a^2}.$$ Substitute $(1,0)$: $$\frac{-2a}{-2(a+b)}=\frac{1-a^2}{b^2-a^2}.$$ So $$\frac{a}{a+b}=\frac{1-a^2}{(b-a)(a+b)}.$$ Multiplying by $(a+b)$, $$a=\frac{1-a^2}{b-a}.$$ Hence $$a(b-a)=1-a^2$$ $$ab-a^2=1-a^2$$ $$ab=1.$$ So we get the important relation: $$ab=1.$$ --- 3. **Use the length of diagonal $AC$** Given $$AC=\frac{25}{4}.$$ Now $$AC^2=(b^2-a^2)^2+(-2b-2a)^2.$$ That is, $$AC^2=(b^2-a^2)^2+4(a+b)^2.$$ Since $ab=1$, $$b^2-a^2=(b-a)(a+b).$$ Also, $$(b-a)^2=(a+b)^2-4ab=(a+b)^2-4.$$ Let $$s=a+b.$$ Then $$(b^2-a^2)^2=(b-a)^2(a+b)^2=(s^2-4)s^2.$$ Therefore, $$AC^2=(s^2-4)s^2+4s^2=s^4.$$ So $$AC=s^2.$$ Given $AC=\frac{25}{4}$, $$s^2=\frac{25}{4}.$$ Thus $$(a+b)^2=\frac{25}{4}.$$ Taking positive value for the geometric configuration, $$a+b=\frac{5}{2}.$$ --- 4. **Find the area of trapezium $ABCD$** The parallel sides of the trapezium are $AB$ and $CD$ (both are not vertical), while the distance between them is the horizontal distance between the vertical sides $AD$ and $BC$. But the easiest way is to use coordinates directly. Lengths of vertical sides: $$AD=2a-(-2a)=4a,$$ $$BC=2b-(-2b)=4b.$$ Distance between these parallel sides is the difference in $x$-coordinates: $$b^2-a^2=(b-a)(a+b).$$ So area of trapezium is $$\text{Area}=\frac12(AD+BC)\times (b^2-a^2).$$ Hence $$\text{Area}=\frac12(4a+4b)(b^2-a^2)$$ $$=2(a+b)(b^2-a^2).$$ Now $$b^2-a^2=(b-a)(a+b),$$ so $$\text{Area}=2(a+b)^2(b-a).$$ Using $ab=1$ and $(a+b)=\frac52$, $$b-a=\sqrt{(a+b)^2-4ab} =\sqrt{\frac{25}{4}-4} =\sqrt{\frac{9}{4}}=\frac32.$$ Therefore, $$\text{Area}=2\cdot \frac{25}{4}\cdot \frac32 =\frac{75}{4}.$$ --- 5. **Check options** The area is $$\boxed{\frac{75}{4}}.$$ So the correct option is: **D. $\frac{75}{4}$**

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View written solutionFree

Correct answer: D

Parametrize points on the parabola

For the parabola $y^2=4x$ , a general point can be written as $P(t)=(t^2,2t).$

Since $AD$ and $BC$ are parallel to the $y$ -axis, points on each of these sides must have the same $x$ -coordinate.

So take:

$A=(a^2,2a)$ and $D=(a^2,-2a)$ ,
$B=(b^2,2b)$ and $C=(b^2,-2b)$ .

Thus $AD$ and $BC$ are vertical chords.

Use the fact that diagonal $AC$ passes through $(1,0)$

Points $A$ and $C$ are:

\qquad C=(b^2,-2b).$$ Since $(1,0)$ lies on line $AC$, the slope from $A$ to $(1,0)$ equals the slope from $(1,0)$ to $C$. Using section/collinearity form: Equation of line through $A$ and $C$: $$\frac{y-2a}{-2b-2a}=\frac{x-a^2}{b^2-a^2}.$$ Substitute $(1,0)$: $$\frac{-2a}{-2(a+b)}=\frac{1-a^2}{b^2-a^2}.$$ So $$\frac{a}{a+b}=\frac{1-a^2}{(b-a)(a+b)}.$$ Multiplying by $(a+b)$, $$a=\frac{1-a^2}{b-a}.$$ Hence $$a(b-a)=1-a^2$$ $$ab-a^2=1-a^2$$ $$ab=1.$$ So we get the important relation: $$ab=1.$$ --- 3. **Use the length of diagonal $AC$** Given $$AC=\frac{25}{4}.$$ Now $$AC^2=(b^2-a^2)^2+(-2b-2a)^2.$$ That is, $$AC^2=(b^2-a^2)^2+4(a+b)^2.$$ Since $ab=1$, $$b^2-a^2=(b-a)(a+b).$$ Also, $$(b-a)^2=(a+b)^2-4ab=(a+b)^2-4.$$ Let $$s=a+b.$$ Then $$(b^2-a^2)^2=(b-a)^2(a+b)^2=(s^2-4)s^2.$$ Therefore, $$AC^2=(s^2-4)s^2+4s^2=s^4.$$ So $$AC=s^2.$$ Given $AC=\frac{25}{4}$, $$s^2=\frac{25}{4}.$$ Thus $$(a+b)^2=\frac{25}{4}.$$ Taking positive value for the geometric configuration, $$a+b=\frac{5}{2}.$$ --- 4. **Find the area of trapezium $ABCD$** The parallel sides of the trapezium are $AB$ and $CD$ (both are not vertical), while the distance between them is the horizontal distance between the vertical sides $AD$ and $BC$. But the easiest way is to use coordinates directly. Lengths of vertical sides: $$AD=2a-(-2a)=4a,$$ $$BC=2b-(-2b)=4b.$$ Distance between these parallel sides is the difference in $x$-coordinates: $$b^2-a^2=(b-a)(a+b).$$ So area of trapezium is $$\text{Area}=\frac12(AD+BC)\times (b^2-a^2).$$ Hence $$\text{Area}=\frac12(4a+4b)(b^2-a^2)$$ $$=2(a+b)(b^2-a^2).$$ Now $$b^2-a^2=(b-a)(a+b),$$ so $$\text{Area}=2(a+b)^2(b-a).$$ Using $ab=1$ and $(a+b)=\frac52$, $$b-a=\sqrt{(a+b)^2-4ab} =\sqrt{\frac{25}{4}-4} =\sqrt{\frac{9}{4}}=\frac32.$$ Therefore, $$\text{Area}=2\cdot \frac{25}{4}\cdot \frac32 =\frac{75}{4}.$$ --- 5. **Check options** The area is $$\boxed{\frac{75}{4}}.$$ So the correct option is: **D. $\frac{75}{4}$**

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