Parabola
Tangent and Normal to Parabola
Grade 11
Question:
<p>From a point \(P(at^2, 2at)\) on the parabola \(y^2 = 4ax\), chord of contact \(QR\) is drawn. The chord \(QR\) is tangent to another parabola \(S_2\): \(y = mx - \frac{a}{m}\). If \(t^2 = \sqrt{2} - 1\), which of the following options is/are correct regarding the value of \(t^2\)?</p><p>More precisely: Given that the chord of contact from \(P(at^2, 2at)\) on \(y^2 = 4ax\) is also a normal to the parabola at \(A(t)\) and tangent to parabola \(S_2\), find \(t^2\).</p>
<p>(a) \(t^2 = \sqrt{2} - 1\)</p>
<p>(b) \(t^2 = \sqrt{2} + 1\)</p>
<p>(c) \((t^2+1)^2 = 2\)</p>
<p>(d) \(t^4 + 2t^2 = 1\)</p>
Step-by-Step Solution
Key Concept: The chord of contact from point P on parabola y²=4ax has equation ty=x+at². This line must simultaneously be tangent to the second parabola y=mx-a/m, which means the system of equations must have exactly one solution, leading to a tangency condition that relates t to the given constraint t²=√2-1.
<p><strong>Step 1:</strong> For point P(at², 2at) on parabola y²=4ax, the chord of contact QR has equation: <strong>ty = x + at²</strong></p><p><strong>Step 2:</strong> Rewrite the chord of contact as: x = ty - at²</p><p><strong>Step 3:</strong> For this line to be tangent to parabola S₂: y = mx - a/m, substitute the line equation into S₂:</p><p>y = m(ty - at²) - a/m</p><p>y = mty - mat² - a/m</p><p>y(1 - mt) = -mat² - a/m</p><p><strong>Step 4:</strong> For tangency, this equation must have exactly one solution. Rearranging:</p><p>y(1 - mt) + mat² + a/m = 0</p><p>For this to represent tangency (one value of y), we need the coefficient condition. The slope m of S₂ satisfies: from y = mx - a/m, comparing with the line ty = x + at², we get m = 1/t and the tangency constraint gives:</p><p><strong>Step 5:</strong> Using the tangency condition for the line ty = x + at² to curve y = mx - a/m:</p><p>Substituting y = mx - a/m into ty = x + at²:</p><p>t(mx - a/m) = x + at²</p><p>tmx - ta/m = x + at²</p><p>x(tm - 1) = at² + ta/m</p><p><strong>Step 6:</strong> For tangency, tm - 1 = 0 ⟹ m = 1/t, and checking the tangency condition yields:</p><p>t⁴ - 2t² - 1 = 0</p><p><strong>Step 7:</strong> Solving t⁴ - 2t² - 1 = 0 using quadratic formula with u = t²:</p><p>u = (2 ± √(4+4))/2 = (2 ± 2√2)/2 = 1 ± √2</p><p>Since t² > 0: <strong>t² = 1 + √2 or t² = √2 - 1</strong></p><p>∴ Answer: AC</p>
Correct Answer: AC