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  • If the tangents drawn from a point P, to the ellipse <img alt="\mathrm{\frac{x^2}{4}+\frac{y^2}{1}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B4%7D&amp

If the tangents drawn from a point P, to the ellipse \mathrm{\frac{x^2}{4}+\frac{y^2}{1}=1} are mutually perpendicular, find the locus of P.

Option: 1

\mathrm{x^2+y^2=10}


Option: 2

\mathrm{2 x^2+y^2=10}


Option: 3

\mathrm{x^2-y^2=5}


Option: 4

\mathrm{None \, \, of \, \, these}


Answers (1)

best_answer

Let point P be \mathrm{(\alpha, \beta)} then combined equation of tangents \mathrm{\mathrm{PT}_1} and \mathrm{\mathrm{PT}_2} is
\mathrm{ \mathrm{SS}_1=\mathrm{T}^2 \text { where } \mathrm{S}_1=\frac{\alpha^2}{4}+\beta^2-1 }
And \mathrm{T=\frac{x \alpha}{4}+y \beta-1}
\mathrm{ \Rightarrow\left(\frac{x^2}{4}+y^2-1\right) S_1=\left(\frac{x \alpha}{4}+y \beta-1\right)^2 }is equation of

\mathrm{\text { tangents } \mathrm{PT}_1 \text { and } \mathrm{PT}_2 \text {. }}

\mathrm{\Rightarrow \frac{S_1 x^2}{4}+S_1 y^2-S_1=\frac{x^2 \alpha^2}{16}+\beta^2 y^2+1+\frac{2 \alpha \beta}{4} x y-2 \beta y-\frac{\alpha}{2} x}

\mathrm{\Rightarrow\left(\frac{S_1}{4}-\frac{\alpha^2}{16}\right) x^2+\left(S_1-\beta^2\right) y^2=S_1+1+\frac{2 \alpha \beta}{4} x y-2 \beta y-\frac{\alpha}{2} x}

Since these are mutually perpendicular so Coefficient of \mathrm{x^2+} Coefficient of \mathrm{y^2=0}

\mathrm{\Rightarrow \frac{S_1}{4}-\frac{\alpha^2}{16}+S_1-\beta^2=0 \Rightarrow \frac{5 S_1}{4}-\frac{\alpha^2}{16}-\beta^2=0}

\mathrm{\Rightarrow \frac{5}{4}\left(\frac{\alpha^2}{4}+\beta^2-1\right)-\frac{\alpha^2}{16}-\beta^2=0 \Rightarrow\left(\frac{5}{16}-\frac{1}{16}\right) \alpha^2+\left(\frac{5}{4}-1\right) \beta^2-\frac{5}{4}=0}

\mathrm{\Rightarrow \frac{\alpha^2}{4}+\frac{\beta^2}{4}-\frac{5}{4}=0 \Rightarrow \alpha^2+\beta^2=5 \text { or } x^2+y^2=5 \text { is the locus of point ' } P \text { '. }}

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vinayak

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