Parabola
Grade 11

Question:

<p>The locus of the middle points of the focal chords of the parabola, y<sup>2</sup> = 4x is:</p>
<p style="display:inline">y<sup>2</sup> = 2(1 - x)</p>
<p style="display:inline">y<sup>2</sup> = 3(x - 1)</p>
<p style="display:inline">y<sup>2</sup> = x - 1</p>
<p style="display:inline">y<sup>2</sup> = 2(x - 1)</p>

Step-by-Step Solution

Key Concept: Represent the midpoint's coordinates using the parametric endpoints of a focal chord and eliminate the parameter $t$ using the algebraic relationship between $(t-1/t)$ and $(t^2+1/t^2)$.
<p>We have&nbsp;<br /> y<sup>2&nbsp;</sup>= 4x<br /> We know that ends of focal chords are&nbsp;(at<sup>2</sup>, 2at)&nbsp;and&nbsp;<span class="math-tex">\(\left(\frac{a}{t^{2}},-\frac{2 a}{t}\right)\)</span><br /> here a = 1<br /> Let (h, k) be the mid point of the chord.<br /> <span class="math-tex">\(\Rightarrow {k}=\frac{2 {t}+\left(-\frac{2}{t}\right)}{2}\)</span><br /> <span class="math-tex">\(\Rightarrow 2 {k}=2 {t}-\frac{2}{{t}}\)</span><br /> <span class="math-tex">\(\Rightarrow {k}={t}-\frac{{1}}{{t}}\)</span>&nbsp;...(i)<br /> h =&nbsp;<span class="math-tex">\(\frac{\mathbf{t}^{2}+\frac{{1}}{{t}^{2}}}{{2}}\)</span><br /> <span class="math-tex">\(\Rightarrow 2 h=\left(t-\frac{1}{t}\right)^{2}\)</span>&nbsp;+ 2<br /> <span class="math-tex">\(\Rightarrow\)</span>&nbsp;2h = k<sup>2</sup> + 2<br /> To get equation of locus replace<br /> h <span class="math-tex">\(\rightarrow\)</span>&nbsp;x&nbsp;and&nbsp;k <span class="math-tex">\(\rightarrow\)</span>&nbsp;y<br /> 2x = y<sup>2</sup> + 2<br /> y<sup>2&nbsp;</sup>= 2(x &minus; 1)</p>
Correct Answer: D

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