Question:
<p>If the normal to y<sup>2</sup> = 12x at P(3, 6) meets the parabola again at a point Q. The circle on the normal chord PQ as diameter is given by</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 30x + 12y - 27 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 30x + 12y - 27 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 30x - 12y - 27 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 30x + 12y + 27 = 0</p>
Step-by-Step Solution
Key Concept: Find the coordinates of the second endpoint Q using the normal chord parametric relationship t2 = -t1 - 2/t1 and apply the diameter form to determine the circle equation.
<p>a = 3, <span class="math-tex">$at_{1}^{2}$</span> = 3<br />
<span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$t_{1}^{2}$</span> = 1 and 2at<sub>1</sub> = 6 <span class="math-tex">$\Rightarrow$</span> t<sub>1</sub> = 1<br />
Let point Q = (<span class="math-tex">$a t_{2}^{2}$</span> , 2at<sub>2</sub>)<br />
Where<br />
t<sub>2</sub> = t<sub>1</sub> - <span class="math-tex">$\frac{2}{\mathrm{t}_{1}}$</span><br />
<span class="math-tex">$\Rightarrow$</span> t<sub>2</sub> = -3<br />
Q = (27, -18)<br />
The equation of the circle on PQ as a diameter..<br />
(x - 3)(x - 27) + (y - 6)(y + 18) = 0<br />
<span class="math-tex">$\Rightarrow$</span> x<sup>2</sup> + y<sup>2</sup> - 30x + 12y - 27 = 0</p>
Correct Answer: A