Question:
<p>The tangent to the parabola y<sup>2</sup> = 4x at the point where it intersects the circle x<sup>2</sup> + y<sup>2</sup> = 5 in the first quadrant, passes through the point:</p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{3}{4}, \frac{7}{4}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{1}{3}, \frac{4}{3}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(\frac{1}{4}, \frac{3}{4}\right)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\left(-\frac{1}{4}, \frac{1}{2}\right)\)</span></p>
Step-by-Step Solution
Key Concept: Find the point of intersection in the first quadrant by substituting the parabola's equation into the circle's equation and then apply the $T=0$ point-form tangent formula $yy_1 = 2a(x+x_1)$.
<p>To find intersection point of x<sup>2</sup> + y<sup>2</sup> = 5 and y<sup>2</sup> = 4x,<br />
substitute y<sup>2</sup> = 4x in x<sup>2</sup> + y<sup>2</sup> = 5, we get,<br />
x<sup>2</sup> + 4x - 5 = 0 <span class="math-tex">$\Rightarrow$</span> x<sup>2</sup> + 5x - x - 5 = 0<br />
<span class="math-tex">$\Rightarrow$</span> x(x + 5) - 1 (x + 5) = 0<br />
<span class="math-tex">$\therefore$</span> x = 1, -5<br />
Intersection point in 1st quadrant be (1, 2)<br />
Now, equation of tangent to y<sup>2</sup> = 4x at (1, 2) is y <span class="math-tex">$\times$</span> 2 = 2(x + 1) <span class="math-tex">$\Rightarrow$</span> y = x + 1<br />
<span class="math-tex">$\Rightarrow$</span> x - y + 1 = 0 ...(i)<br />
Hence, <span class="math-tex">$\left(\frac{3}{4}, \frac{7}{4}\right)$</span> lies on (i)</p>
Correct Answer: A