Question:
<p>The tangents from a point P to y<sup>2</sup> = 4ax have the inclinations <span class="math-tex">\(\theta_{1}\)</span> and <span class="math-tex">\(\theta_{2}\)</span>, and cot <span class="math-tex">\(\theta_{1}\)</span> + cot <span class="math-tex">\(\theta_{2}\)</span> = 2, then point P lies on</p>
<p style="display:inline">x = 2a</p>
<p style="display:inline">x + 2a = 0</p>
<p style="display:inline">y + 2a = 0</p>
<p style="display:inline">y = 2a</p>
Step-by-Step Solution
Key Concept: For a parabola y² = 4ax, if tangents from external point P(h,k) touch at points with parameters t₁ and t₂, then the slopes are m₁ = 1/t₁ and m₂ = 1/t₂. Using cot θ₁ + cot θ₂ = 2 translates to t₁ + t₂ = 2, which gives the y-coordinate of P as k = a(t₁ + t₂) = 2a.
<p>Let the tangents from point P meet the curve at Q and R.<br />
Let Q = (at<sub>1</sub><sup>2</sup>, 2at<sub>1</sub>) and R = (at<sub>2</sub><sup>2</sup>, 2at<sub>2</sub>).<br />
Then P = (at<sub>1</sub>t<sub>2</sub>, a(t<sub>1</sub> + t<sub>2</sub>))<br />
The slopes of the tangents at Q and R are <span class="math-tex">$\frac{1}{t_{1}}$</span> and <span class="math-tex">$\frac{1}{t_{2}}$</span> respectively.<br />
<span class="math-tex">$\Rightarrow \tan \theta_{1}=\frac{1}{t_{1}}, \tan \theta_{2}=\frac{1}{t_{2}}$</span><br />
Given cot cot <span class="math-tex">$\theta_1$</span> + cot<span class="math-tex">$\theta_2$</span> = 2<br />
<span class="math-tex">$\Leftrightarrow$</span> t<sub>1</sub> + t<sub>2</sub> = 2<br />
Ordinate of point P is a(t<sub>1</sub> + t<sub>2</sub>) = 2a<br />
<span class="math-tex">$\Rightarrow$</span> Point P lies on the line y = 2a.</p>
Correct Answer: D