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Hyperbola

Section: Mathematics

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Question : 4 of 34

Marks: +1, -0

If

3x+2\sqrt{2}y+k=0

is a normal to the hyperbola

4x^2-9y^2-36=0

making positive intercepts on both the axes, then

k=

[AP EAPCET 23 May 2025 S2]

$13\sqrt{2}$
$-5\sqrt{2}$
$-2\sqrt{2}$
$-13\sqrt{2}$

Solution:

4x^2-9y^2=36 \Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=0

So,

a^2=9, b^2=4

Thus, equation of normal

\frac{9x}{x_1}+\frac{4y}{y_1}=(9+4)=13 \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

and the given equation is

3x+2\sqrt{2}y+k=0

\Rightarrow \;\; 3x+2\sqrt{2}y=-k \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

∵ Both equation are proportional

Thus,

\frac{\frac{9}{x_1}}{3}=\frac{\frac{4}{y_1}}{2\sqrt{2}}=\frac{13}{-k}

\Rightarrow \;\; \frac{3}{x_1}=\frac{\sqrt{2}}{y_1} \Rightarrow x_1=\frac{3}{\sqrt{2}} y_1

Put in equation of hyperbola

\frac{\left(\frac{3}{\sqrt{2}} y_1\right)^2}{9}-\frac{y^2}{4}=1 \Rightarrow y_1^2=4 \Rightarrow y_1= \pm 2

at

y_1=2 \Rightarrow x_1=3\sqrt{2}

and at

y_1=-2

\Rightarrow x_1=-3\sqrt{2}

Also,

\frac{3}{x_1}=-\frac{13}{k} \Rightarrow -k=\frac{13}{3} \times 3\sqrt{2}=13\sqrt{2}

Hence,

k=-13\sqrt{2}

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