If 4{x^2} + p{y^2} = 45 and {x^2} - 4{y^2} = 5 cut orthogonally, then value of p is

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10-2. Parabola, Ellipse, Hyperbola

hard

If $4{x^2} + p{y^2} = 45$ and ${x^2} - 4{y^2} = 5$ cut orthogonally, then the value of $p$ is

A

$1\over9$

B

$1\over3$

C

$3$

D

$9$

Solution

(D) Slope of $1^{st}$ curve ${\left( {\frac{{dy}}{{dx}}} \right)_I} = – \frac{{4x}}{{py}}$

Slope of $2^{nd}$ curve ${\left( {\frac{{dy}}{{dx}}} \right)_{II}} = \frac{x}{{4y}}$

For orthogonal intersection $\left( { – \frac{{4x}}{{py}}} \right)\,\left( {\frac{x}{{4y}}} \right) = – 1$

==> ${x^2} = p{y^2}$

On solving equations of given curves $x = 3$, $y = 1$

$p(1) = {(3)^2} = 9$

==> $p = 9$.

Standard 11

Mathematics

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