Straight line x + y = √ 2 p will touch hyperbola 4{x^2} - 9{y^2} = 36 , if

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10-2. Parabola, Ellipse, Hyperbola

easy

The straight line $x + y = \sqrt 2 p$ will touch the hyperbola $4{x^2} - 9{y^2} = 36$, if

A

${p^2} = 2$

B

${p^2} = 5$

C

$5{p^2} = 2$

D

$2{p^2} = 5$

Solution

(d) The condition for the line $y = mx + c$ will touch the hyperbola

$\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ is ${c^2} = {a^2}{m^2}$$ – {b^2}$

Here $m = – 1$, $c = \sqrt 2 p,$ ${a^2} = 9,\,\,{b^2} = 4$

We get $2{p^2} = 5.$

Standard 11

Mathematics

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