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Equation of pair of tangents to the ellipse \mathrm{9 x^2+25 y^2=225} from a point (4,2) is

Option: 1

\mathrm{9 x^2+25 y^2-225=\left(\frac{36 x+50 y-225}{\sqrt{19}}\right)^2}


Option: 2

\mathrm{9 x^2+25 y^2-225=\left(\frac{36 x+25 y-225}{15 \sqrt{19}}\right)^2}


Option: 3

\mathrm{19\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)=\left(\frac{36 x+25 y-225}{15}\right)^2}


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\text { Equation of pair of tangents to an ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

\mathrm{\text { at }\left(x_1, y_1\right) \text { is given by }}

\mathrm{\begin{aligned} & \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)=\left(\frac{x x_1}{a^2}+\frac{y y_1}{b^2}-1\right)^2 \\ & \text { or } S S^{\prime}=T^2 \end{aligned}}

\mathrm{\text { Here, }\left(x_1, y_1\right)=(4,2) \text {, }}

\mathrm{\begin{aligned} T & =\left(\frac{x x_1}{a^2}+\frac{y y_1}{b^2}-1\right)=\left(\frac{4 x}{25}+\frac{2 y}{9}-1\right), \\ S & =\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=\frac{x^2}{25}+\frac{y^2}{9}-1 \\ \text { and } S^{\prime} & =\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1=\left(\frac{16}{25}+\frac{4}{9}-1\right) \end{aligned}}

Required equation of pair of tangents is \mathrm{S S^{\prime}=T^2}

\mathrm{\begin{aligned} & \Rightarrow\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)\left(\frac{16}{25}+\frac{4}{9}-1\right)=\left(\frac{4 x}{25}+\frac{2 y}{9}-1\right)^2 \\ & \text { or, }\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)=\left(\frac{36 x+50 y-225}{15 \sqrt{19}}\right)^2 \end{aligned}}

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Ritika Harsh

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