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  • The midpoint of chord of contact of the point (5,1) to the hyperbola <img alt="\mathrm{x^2-4 y^2=16}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2-4%20y%5E2%3D16%7D"

The midpoint of chord of contact of the point (5,1) to the hyperbola \mathrm{x^2-4 y^2=16} is

Option: 1

\left(\frac{16}{21}, \frac{80}{21}\right)


Option: 2

\left(\frac{80}{21}, \frac{16}{21}\right)


Option: 3

\left(\frac{60}{22}, \frac{16}{22}\right)


Option: 4

\left(\frac{16}{22}, \frac{60}{22}\right)


Answers (1)

best_answer

The equation of the chord of contact of (5,1) is

\mathrm{ \begin{aligned} & \frac{5 x}{16}-\frac{y}{4}=1 . \\ & \Rightarrow 5 x-4 y=16 . \end{aligned} }                           ...(1)

If \mathrm{(\alpha, \beta)} be the mid point of the chord then,

\mathrm{ \frac{\alpha x}{16}-\frac{\beta y}{4}-1=\frac{\alpha^2}{16}-\frac{\beta^2}{4}-1 }                ...(2)

From (1) and (2) \mathrm{\alpha=\frac{80}{21}, \beta=\frac{16}{21}}

Hence, mid point is \mathrm{\left(\frac{80}{21}, \frac{16}{21}\right).}

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Gunjita

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