The equation of the common tangent to the curves y² = 16x and xy = - 4 is (a) x-y+4=0 (b) x+y+4=0 (c) x-2y+16=0 (d) 2x-y+2=0

Answer: (a) x - y + 4 = 0 Any tangent to the parabola y² = 16x (here 4a = 16, so a = 4) with slope m is y = mx + 4/m For it to touch the hyperbola xy = -4, substitute for y: x(mx + 4/m) = -4 mx² + 4/mx + 4 = 0 A tangent meets the curve at exactly one point, so the discriminant must be zero: frac16m² - 16m = 0 ⇒ m³ = 1 ⇒ m = 1 So the common tangent is y = x + 4, i.e. x - y + 4 = 0. Check: it touches y²=16x at (4, 8) and xy=-4 at (-2, 2).

Chapter: Uncategorized· Mathematics

The equation of the common tangent to the curves y^2 = 16x and xy = - 4 is

The equation of the common tangent to the curves y² = 16x and xy = - 4 is

(a) x-y+4=0
(b) x+y+4=0
(c) x-2y+16=0
(d) 2x-y+2=0

Doubtly Asked 3 Jun 2021 Updated 23 Oct 2026 Student doubt jeemain Save WhatsApp Report

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Answer: (a) \(x - y + 4 = 0\)

Any tangent to the parabola \(y^{2} = 16x\) (here \(4a = 16\), so \(a = 4\)) with slope \(m\) is

\[y = mx + \frac{4}{m}\]

For it to touch the hyperbola \(xy = -4\), substitute for \(y\):

\[x\left(mx + \frac{4}{m}\right) = -4\]

\[mx^{2} + \frac{4}{m}x + 4 = 0\]

A tangent meets the curve at exactly one point, so the discriminant must be zero:

\[\frac{16}{m^{2}} - 16m = 0 \;\Rightarrow\; m^{3} = 1 \;\Rightarrow\; m = 1\]

So the common tangent is \(y = x + 4\), i.e. \(x - y + 4 = 0\).

Check: it touches \(y^{2}=16x\) at \((4, 8)\) and \(xy=-4\) at \((-2, 2)\).

Doubtly AI · 23 Oct 2026 AI-verified · Oct 2026 Report

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