Question:
<p>The tangents to the curve y = (x - 2)<sup>2</sup> - 1 at its points of intersection with the line x - y = 3, intersect at the point</p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{5}{2},-1\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{5}{2}, 1\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{5}{2},-1\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{5}{2},1\right)\)</span></p>
Step-by-Step Solution
Key Concept: The intersection point of the tangents is the pole of the given line, which is solved by comparing the line's equation with the parabola's chord of contact formula T=0.
<p>Given equation of parabola is<br />
y = (x - 2)<sup>2</sup> - 1<br />
<span class="math-tex">\(\Rightarrow\)</span> y = x<sup>2</sup> - 4x + 3 ... (i)<br />
Now, let (x<sub>1</sub>, y<sub>1</sub>) be the point of intersection of tangents of parabola (i) and line x - y = 3, then<br />
Equation of chord of contact of point (x<sub>1</sub>, y<sub>1</sub>) w.r.t. parabola (i) is<br />
T = 0<br />
<span class="math-tex">\(\Rightarrow \frac{1}{2}\left(y+. y_{1}\right)=x x_{1}-2\left(x+x_{1}\right)+3\)</span><br />
<span class="math-tex">\(\Rightarrow y+y_{1}=2 x\left(x_{1}-2\right)-4 x_{1}+6\)</span><br />
<span class="math-tex">\(\Rightarrow 2 x\left(x_{1}-2\right)-y=4 x_{1}+y_{1}-6\)</span>, this equation represent the line x - y = 3 only, so on comparing, we get<br />
<span class="math-tex">\(\frac{2\left(x_{1}-2\right)}{1}=\frac{-1}{-1}=\frac{4 x_{1}+y_{1}-6}{3}\)</span><br />
<span class="math-tex">\(\Rightarrow x_{1}=\frac{5}{2}\)</span> and y<sub>1</sub> = -1<br />
So, the required point is <span class="math-tex">\(\left(\frac{5}{2},-1\right)\)</span>.</p>
Correct Answer: A