N2022 P1 Q5

N2022 P1 Q5

Junior College 2
8 marks
Free

The line with equation \(y=mx\) is a tangent to the curve with equation

\[(x+8)^2+(y-14)^2=52.\]

  1. Show that \(m\) satisfies the equation \(3m^2+56m+36=0\).[4]

\(A\) and \(B\) are points on the curve. The tangent at \(A\) and the tangent at \(B\) intersect at the origin.

  1. Find the coordinates of \(A\) and \(B\).[4]

Default solution

  1. \(y=mx\Rightarrow (x+8)^2+(mx-14)^2=52\)
    \((m^2+1)x^2+(16-28m)x+208=0\)
    \(\text{Tangency}\Rightarrow (16-28m)^2-4(m^2+1)(208)=0\)
    \(\therefore\ 3m^2+56m+36=0\) (shown)
  2. \(3m^2+56m+36=0\Rightarrow m=-\frac23\text{ or}\hspace{0.5em}m=-18\)
    \(m=-\frac23\Rightarrow x=-12,\ y=8\)
    \(m=-18\Rightarrow x=-\frac45,\ y=\frac{72}{5}\)
    \(A,B:\ (-12,8)\text{ and}\hspace{0.5em}\left(-\frac45,\frac{72}{5}\right)\) (either order)
Answer:(a) \(3m^2+56m+36=0\) (b) \((-12,8),\ \left(-\frac45,\frac{72}{5}\right)\) (either order)

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