<p>Let ABCD be a trapezium whose vertices lie on the parabola <span class="math-tex">\(y^{2}=4 x\)</span>. Let the sides AD and BC of the trapezium be parallel to <span class="math-tex">\(y\)</span>-axis. If the diagonal <span class="math-tex">\(A C\)</span> is of length <span class="math-tex">\(\frac{25}{4}\)</span> and it passes through the point <span class="math-tex">\((1,0)\)</span>, then the area of ABCD is</p>
<p style="display:inline"><span class="math-tex">\(\frac{125}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{75}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{75}{8}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{25}{2}\)</span></p>
Step-by-Step Solution
Key Concept: Recognize that diagonal AC is a focal chord because it passes through the focus (1,0), allowing the use of the focal chord length formula $a(t+1/t)^2$ to find the vertices and parallel side lengths.
<p><img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775822924-wma329.jpg" style="height:157px; width:200px" /><br />
Given, <span class="math-tex">$y^{2}=4 x$</span><br />
Also AD & BC are parallel to y -axis Length of diagonal <span class="math-tex">${AC}=\frac{25}{4}$</span>, also <span class="math-tex">$(1,0)$</span> lie on diagonal AC Now, as shown in figure,<br />
<span class="math-tex">$A\left(a t_{1}^{2}, 2 a t_{1}\right), {C}\left(\frac{a}{t_{1}^{2}},-\frac{2 a}{t_{1}}\right)$</span><br />
Now, applying distance formula between A and C, we get<br />
<span class="math-tex">${AC}=\left(t_{1}+\frac{1}{t_{1}}\right)^{2}=\frac{25}{4}$</span>,<br />
<span class="math-tex">$\Rightarrow t_{1}+\frac{1}{t_{1}}= \pm \frac{5}{2}$</span><br />
On simplifying, we get<br />
<span class="math-tex">$\Rightarrow t=2$</span> or <span class="math-tex">$\frac{1}{2}$</span><br />
<span class="math-tex">$\therefore {A}\left(\frac{1}{2}, 1\right), {D}\left(\frac{1}{4},-1\right), {B}(4,4), {C}(4,-4)$</span><br />
Now, area of trapezium<br />
<span class="math-tex">$=\frac{1}{2}$</span> (sum of diagonals) (height)<br />
<span class="math-tex">$=\frac{1}{2}(8+2)\left(4-\frac{1}{4}\right)=\frac{75}{4}$</span> sq. units</p>
Correct Answer: B