Question

The equations of the tangents to the ellipse 9x² + 16y² = 144 from the point (2, 3) are
(a) y = 3, x = 5
(b) x = 2, y = 3
(c) x = 3, y = 2
(d) x + y = 5, y = 3

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)x+y=5,y=3 \\ 9x^2+16y^2=144 \\ \Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1 \\ \text{Equation of the tangent in case of an ellipse is given by} \\ y=mx+\sqrt{a^2{m}^2+b^2} \\ \Rightarrow y=mx+\sqrt{16m^2+9}...\left(1\right) \\ \text{Substituting}\mathit{\text{x}}\text{=2 and}\mathit{\text{y}}\text{=3, we get:} \\ 3=2m\pm \sqrt{16m^2+9} \\ \Rightarrow 3-2m=\sqrt{16m^2+9} \\ \text{On squaring both sides, we get:} \\ {\left(3-2m\right)}^2=\left(16m^2+9\right) \\ \Rightarrow 9+4m^2-12m=\left(16m^2+9\right) \\ \Rightarrow 12m^2+12m=0 \\ \Rightarrow 12m\left(m+1\right)=0 \\ \Rightarrow m=0,-1 \\ \text{Substituting values of}\mathit{\text{m}}\text{in eq. (1), we get:} \\ \text{For}m=0,y=3 \\ \mathrm{For}m=-1,y=-x+5\mathrm{or}x+y=5 \end{gathered}$$