Question:
<p>Let <span class="math-tex">\(P\)</span> be the parabola, whose focus is <span class="math-tex">\((-2,1)\)</span> and directrix is <span class="math-tex">\(2 x+ y+2=0\)</span>. Then the sum of the ordinates of the points on <span class="math-tex">\(P\)</span>, whose abscissa is -2, is</p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{5}{2}\)</span></p>
Step-by-Step Solution
Key Concept: The fundamental definition of a parabola states that the distance from any point on the curve to the focus is equal to its perpendicular distance to the directrix (PS = PM).
<p>For any parabola, the distance from a point <span class="math-tex">$(x, y)$</span> to the focus is equal to its perpendicular distance to the directrix. Using this definition, we write:<br />
<span class="math-tex">$\sqrt{(x+2)^{2}+(y-1)^{2}}=\frac{|2 x+y+2|}{\sqrt{5}}$</span>.<br />
<span class="math-tex">$5\left[(x+2)^{2}+(y-1)^{2}\right]=(2 x+y+2)^{2}$</span><br />
<img src="https://media-mycbseguide.s3.amazonaws.com/images/question_images/1775822032-mxa734.jpg" style="height:167px; width:200px" /><br />
Substituting <span class="math-tex">$x=-2$</span> into the equation:<br />
<span class="math-tex">$5\left(0+(y-1)^{2}\right)=(y-2)^{2}$</span><br />
<span class="math-tex">$\Rightarrow 5\left(y^{2}-2 y+1\right)=y^{2}-4 y+4$</span><br />
<span class="math-tex">$\Rightarrow 5 y^{2}-10 y+5=y^{2}-4 y+4$</span><br />
<span class="math-tex">$\Rightarrow 4 y^{2}-6 y+1=0$</span><br />
For a quadratic equation <span class="math-tex">$a y^{2}+b y+c =0$</span>, the sum of the roots is given by<br />
<span class="math-tex">$y_{1}+y_{2}=-\frac{b}{a}$</span>.<br />
<span class="math-tex">$y_{1}+y_{2}=\frac{6}{4}=\frac{3}{2}$</span>.</p>
Correct Answer: A