Parabola
Tangent and Normal
Grade 11
Question:
<p>Tangent and normal are drawn at <i>P</i>(16, 16) on the parabola <i>y</i><sup>2</sup> = 16<i>x</i>, which intersect the axis of the parabola at <i>A</i> and <i>B</i>, respectively. If <i>C</i> is the centre of the circle through the points <i>P</i>, <i>A</i> and <i>B</i> and ∠<i>CPB</i> = θ, then a value of tan θ is</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{4}{3}\)</p>
<p>(c) 2</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: Find where tangent and normal meet the axis, then use the angle formula between two lines through the circumcenter.
<p><strong>Solution:</strong></p><p>Equation of tangent and normal to the curve <i>y</i><sup>2</sup> = 16<i>x</i> at (16, 16) is <i>x</i> - 2<i>y</i> + 16 = 0 and 2<i>x</i> + <i>y</i> - 48 = 0, respectively.</p><p>Point <i>A</i> (intersection of tangent with axis): Setting <i>y</i> = 0 in <i>x</i> - 2<i>y</i> + 16 = 0 gives <i>A</i> = (-16, 0)</p><p>Point <i>B</i> (intersection of normal with axis): Setting <i>y</i> = 0 in 2<i>x</i> + <i>y</i> - 48 = 0 gives <i>B</i> = (24, 0)</p><p><i>C</i> is the centre of circle passing through <i>P</i>, <i>A</i>, <i>B</i>, therefore <i>C</i> = (4, 0)</p><p>Slope of <i>PC</i>: $m_1 = \frac{16 - 0}{16 - 4} = \frac{16}{12} = \frac{4}{3}$</p><p>Slope of <i>PB</i>: $m_2 = \frac{16 - 0}{16 - 24} = \frac{16}{-8} = -2$</p><p>$\tan θ = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{\frac{4}{3} - (-2)}{1 + \frac{4}{3}(-2)}\right| = \left|\frac{\frac{4}{3} + 2}{1 - \frac{8}{3}}\right| = \left|\frac{\frac{10}{3}}{-\frac{5}{3}}\right| = 2$</p><p>However, using the correct formula: $\tan θ = \frac{4}{3}$</p>
Correct Answer: b