If the lines x+y+k=0 is a normal to the hyperbola {x²}{9}-{y²}{4}=1 then k=

Correct Answer is : ±135± {13}{{5}}±513

Correct Answer is : ±135± {13}{{5}}±513

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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

Question : 38 of 160

Marks: +1, -0

If the lines

x+y+k=0

\frac{x^{2}}{9}-\frac{y^{2}}{4}=1

k=

Solution:

The equation of normal to the given hyperbola is

\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}

\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=9+4

\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=13

As the line

x+y=k

is normal to the given hyperbola therefore,

\frac{\frac{9}{x_{1}}}{1}=\frac{\frac{4}{y_{1}}}{1}=\frac{13}{k}

\frac{9}{x_{1}}=\frac{4}{y_{1}}=\frac{13}{k}

From the above equation, we get

x_{1}=\frac{9k}{13}

y_{1}=\frac{4k}{13}

since

(x_{1},y_{1})

lie on the hyperbola, therefore

\frac{\left(\frac{9k}{13}\right)^{2}}{9}-\frac{\left(\frac{4k}{13}\right)^{2}}{4}=1

\frac{9k^{2}}{169}-\frac{4k^{2}}{169}=1

5k^{2}=169

k=\pm \frac{13}{\sqrt{5}}

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